a) 2³ = 2^x

<=> \(\dfrac{2^3}{2^x}=1\)

<=> 2^{3 - x} = 1

<=> 3 - x = 0

<=> x = 3

b) 25 = 5^x

<=> 5² = 5^x

=> x = 2

c) 32 = 2^x

<=> 2⁴ = 2^x

<=> x = 4

We have :

2³⁰ = (2³)¹⁰ = 8¹⁰ < 9¹⁰ = (3²)¹⁰ = 3²⁰

So 2³⁰ < 3²⁰

@Macedoi , No ..... now i'm not a cheater , and up to now, I have never done.

First thing: Why do you call me a cheater ? 

Be careful your mouth , It is not used to defame others .

Second thing: Do you have evidence ? Do you?

Third thing: This is a spam question , @Phan Thanh Tinh mentioned this , so you don't have to mention again. 

Fourth thing: Hackers, copiers and cheaters , all of you should know that, if you go too far, your nickname will be going away with no goobye ! 

@Macedoi , what about you , do you know :"Tự trả lời tự " in Vietnam ? 

Phan Huy Toàn , phanhuytien , @Help you solve math 

Hey Phan Thanh Tinh , your family right , come and kick it ass off :v 

Ok , compare :))

We have : \(2^{332}< 2^{333}=\left(2^3\right)^{111}< \left(3^2\right)^{111}=3^{222}< 3^{223}\)

That is too difficult , right =)) 

I don't see the picture :V :V :V 

a) x⁴ - x³ + x² - x = 0

<=> x³.(x - 1) + x.(x - 1) = 0

<=> (x³ + x)(x - 1) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x^3+x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\left(x^2+1\right)=0\\x=1\end{matrix}\right.\)

At first case , we can see :

\(x^2+1>0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

b) x² - 8² = 0

<=> (x - 8)(x + 8) = 0

<=> \(\left[{}\begin{matrix}x-8=0\\x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\)

c) (x + 1) + (x + 2) + (x + 3) + ...... + (x + 10) = 55

<=> 10x + (1 + 2 + ...... + 10) = 55

<=> 10x + 45 = 55

<=> 10x = 10

<=> x = 1 

We have the following sequence of natural numbers : 

0 ; 1 ; 2 ; 3 ; ........ ; 522

Let P is the sum of odd numbers which have three-digit natural numbers , we got :

P = 101 ; 103 ; 105 ; ....... ; 521   (99 < P < 522)

Let P' is the sum of odd number which contain number 5 at last  , we got :

P' = 105 ; 115 ; 125 ; ...... ; 515 (99 < P' < 522)

Apply formula to calculate number of numbers in the sequence :

P = \(\dfrac{\left(521-101\right)}{2}+1=\dfrac{420}{2}+1=210+1=211\) (numbers)

P' = \(\dfrac{\left(515-105\right)}{10}+1=\dfrac{410}{10}+1=41+1=42\) (numbers)

So, there are 211 - 42 = 169 odd numbers contain no 5 

a) \(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}+4}+\dfrac{4}{\sqrt{x}-4}\right):\dfrac{x+16}{\sqrt{x}+2}\)

\(B=\dfrac{\sqrt{x}\left(\sqrt{x}-4\right)+4\left(\sqrt{x}+4\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}.\dfrac{\sqrt{x}+2}{x+16}\)

\(B=\dfrac{x-4\sqrt{x}+4\sqrt{x}+16}{x-16}.\dfrac{\sqrt{x}+2}{x+16}\)

\(B=\dfrac{\left(x+16\right).\left(\sqrt{x}+2\right)}{\left(x-16\right)\left(x+16\right)}=\dfrac{\sqrt{x}+2}{x-16}\)

b) B.(A - 1) = \(\dfrac{\sqrt{x}+2}{x-16}.\left(\dfrac{\sqrt{x}+4}{\sqrt{x}+2}-1\right)\)

\(=\dfrac{\sqrt{x}+2}{x-16}.\dfrac{2}{\sqrt{x}+2}=\dfrac{2}{x-16}\)

B(A - 1) is interger number

<=> 2 \(⋮\) x - 16

<=> x - 16 \(\in\) {1 ; -1 ; 2 ; -2}

We have this table :

So, x = {14 ; 15 ; 17 ; 18}

a) We have :

\(\left\{{}\begin{matrix}\widehat{ADH}=90^0\\\widehat{DAE}=90^0\\\widehat{AEH}=90^0\end{matrix}\right.\)

So , quadrilateral DHEA is the rectangle

=> AH = DE (2 diagonals)

It also means : OD = OH = OE = OA (with O is the intersection)

b) Consider \(\Delta ODI\) and \(\Delta OHI\) , we have :

DI = IH (DI is the median of square triagle BDH)

DO = OH

IO general 

=> \(\Delta ODI\) = \(\Delta OHI\) (e - e - e)

=> \(\widehat{IDO}=90^0\Rightarrow DE\perp ID\) (1)

Similar with \(\Delta OHK\) and \(\Delta OEK\) 

\(\Delta OHK\) = \(\Delta OEK\) (e - e - e)

=> \(\widehat{OEK}=90^0\Rightarrow DE\perp EK\) (2)

From (1) and (2) 

=> DI // EK 

Source: Câu hỏi của thảobăng - Toán lớp 9 - Học toán với OnlineMath

I'm just finished ! :v 

Apply Bunyakovsky , we have :

\(\left(1+1+1\right)\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2=\left(\dfrac{3}{2}\right)^2=\dfrac{9}{4}\)

\(\Rightarrow3\left(a^2+b^2+c^2\right)\ge\dfrac{9}{4}\)

\(\Rightarrow a^2+b^2+c^2\ge\dfrac{3}{4}\)

We have the following diagram:

Men:      |----------------------------|

Woman: |-------------| 32(people)

The number of men in group is :

\(\left(212+32\right):2=\dfrac{244}{2}=122\) (people)

So, there are 122 men in the group. 

We have :

\(\dfrac{n^3}{3}+\dfrac{n^3}{2}+\dfrac{n^3}{6}=\dfrac{2n^3}{6}+\dfrac{3n^3}{6}+\dfrac{n^3}{6}\)

\(=\dfrac{2n^3+3n^3+n^3}{6}=\dfrac{6n^3}{6}=n^3\)

So, we have the thing to prove :) 

(2x - 5)(2x + 5) = 5

\(\Leftrightarrow\left(2x\right)^2-5^2=5\)

\(\Leftrightarrow4x^2=5+25=30\)

:v oh..... i'm wrong !!! 

Area on one face is :

\(\dfrac{96}{6}=36\left(cm^2\right)\)

The length of an edge is :

\(\sqrt{36}=6\) (cm)

The volume of the cube is :

\(V=6.6.6=6^3=216\left(cm^3\right)\)

Lê Quốc Trần Anh :v , you wrong 

e) \(\dfrac{x-2}{9}=\dfrac{4}{x-2}\)

\(\Leftrightarrow\left(x-2\right)^2=36\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=6\\x-2=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-4\end{matrix}\right.\)

a) Apply the properties of the ratios are equal , we have :

\(\dfrac{x}{7}=\dfrac{y}{13}=\dfrac{x+y}{7+13}=\dfrac{40}{20}=2\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=14\\y=26\end{matrix}\right.\)

Similar with b,c and d , we have :

b) \(\dfrac{2x}{38}=\dfrac{y}{21}=\dfrac{2x-y}{38-21}=\dfrac{34}{17}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=....\\y=.....\end{matrix}\right.\)

c) It's wrong

d) \(\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{x^2+y^2}{9+16}=\dfrac{100}{25}=4\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=36\\y^2=64\end{matrix}\right.\)

\(\Rightarrow x=\left[{}\begin{matrix}6\\-6\end{matrix}\right.\) and \(y=\left[{}\begin{matrix}8\\-8\end{matrix}\right.\)

a // b

=> \(\widehat{aAB\: }=\widehat{bBA}\) (2 Alternate interior angles) 

Because Al and Bw are bisectors of two angles corresponding .

So \(\widehat{aAl\: }=\widehat{lAB}=\widehat{ABw}=\widehat{wBb}\)

=> \(\widehat{lAB}=\widehat{ABw}\)

But two angles are 2 Alternate interior angles

=> Al // Bw