|x+3|+|x+5|-|x+1/2|+|7-x|

TH1: -x-3-x-5+x-1/2+7-x

    = -2x-3/2

TH2: -x-3-x-5+x-1/2+x-7

   = 0-31/2

=-31/2

Câu 1:

Ta có \(x^3+3x-5=x^3+2x+x-5=\left(x^2+2\right)x+x-5\)

để giá trị của đa thức \(x^3+3x-5\)chia hết cho giá trị của đa thức \(x^2+2\)

thì \(x-5⋮x^2+2\Rightarrow\left(x-5\right)\left(x+5\right)⋮x^2+2\Rightarrow x^2-25⋮x^2+2\)

\(\Leftrightarrow x^2+2-27⋮x^2+2\Rightarrow27⋮x^2+2\)

\(\Leftrightarrow x^2+2\inƯ\left(27\right)\)do \(x^2+2\inℤ,\forall x\inℤ\)

mà \(x^2+2\ge2,\forall x\inℤ\)

\(\Rightarrow x^2+2\in\left\{3;9;27\right\}\)\(\Leftrightarrow x^2\in\left\{1;7;25\right\}\)

mà \(x^2\)là số chính phương \(\forall x\inℤ\)

\(\Rightarrow x^2\in\left\{1;25\right\}\Leftrightarrow x\in\left\{\pm1;\pm5\right\}\)

**bạn nhớ thử lại nhé
\(KL...\)

Bạn Minh Tâm ơi giá trị \(\pm1\)sai rồi

2x²+2y²=5xy

<=>2x²-5xy+2y²=0

<=>(2x²-4xy)-(xy-2y²)=0

<=>2x(x-2y)-y(x-2y)=0

<=>(x-2y).(2x-y)=0

<=> (x-2y)=0 hoặc 2x-y=0

Nếu x-2y=0 =>x=2y

=>E=\(\frac{x+y}{x-y}\)=\(\frac{2y+y}{2y-y}\)=\(\frac{3y}{y}\)=3

Nếu 2x-y=0 =>2x=y

=>E=\(\frac{x+y}{x-y}\)=\(\frac{x+2x}{x-2x}\)=\(\frac{3x}{-1x}\)= -3

2x^2 + 2y^2 = 5xy

<=> 2x^2 + 2y^2 - 5xy = 0

<=> 2x^2  - 4xy + 2y^2 - xy  = 0

<=> 2x(x - 2y) - y(x - 2y) = 0

<=> (2x - y)(x - 2y) = 0

<=> 2x = y hoặc x = 2y

thay vào là xong

Ta có :\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\)

\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}\right)^3=-\frac{1}{c^3}\)

\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=-3\cdot\frac{1}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)=3\cdot\frac{1}{abc}\)

( Do \(\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\) )

Khi đó : \(P=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=abc\cdot\frac{3}{abc}=3\)

\(5+4x-x+2=\left(5x+4\right)\left(7+5x\right)\)

\(\Leftrightarrow5+4x-x+2=35+28x+25x+20x^2\)

\(\Leftrightarrow x^2+50x+28=0\)

Ta có \(\Delta=50^2-4.1.28=2388,\sqrt{\Delta}=2\sqrt{597}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-50+2\sqrt{597}}{2}=-25+\sqrt{597}\\x=\frac{-50-2\sqrt{597}}{2}=-25-\sqrt{597}\end{cases}}\)

\(5+4x-x+2=\left(5+4x\right)\left(7+5x\right)\)

\(7+3x=\left(5+4x\right)\left(7+5x\right)\)

\(7+3x=35+28x+25x+20x^2\)

\(7+3x-35-28x-25x-20x^2=0\)

\(-28-50x-20x^2=0\)

\(-28-50x-20x^2=0\)

\(x=-\frac{25+\sqrt{65}}{20};-\frac{25-\sqrt{65}}{20}\)

Ta có : \(x^2+5y^2+2x-4xy-10y+14\)

\(=x^2+2x\left(1-2y\right)+\left(1-2y\right)^2-\left(1-2y\right)^2+5y^2-10y+14\)

\(=\left(x-2x+1\right)^2-1-4y^2+4y+5y^2-10y+14\)

\(=\left(x-2x+1\right)^2+y^2-6y+9+4\)

\(=\left(x-2x+1\right)^2+\left(y-3\right)^2+4\ge4>0\) (đpcm)

Ta có: x² + 5y² + 2x - 4xy - 10y + 14 

= (x² - 4xy + 4y²) + (2x - 4y) + 1 + (y² - 6y + 9) + 4

= (x - 2y)² + 2(x - 2y) + 1 + (y - 3)² + 4

= (x - 2y + 1)² + (y - 3)² + 4 > 0 \(\forall\)x; y

Do (x - 2y + 1)² \(\ge\)0; (y - 3)² \(\ge\)0 ; 4 > 0

a) \(pt\Leftrightarrow\frac{6}{x^2+2}-1+\frac{7}{x^2+3}-1+\frac{12}{x^2+8}-1-\frac{3x^2+16}{x^2+10}+2=0\)

\(\Leftrightarrow\frac{4-x^2}{x^2+2}+\frac{4-x^2}{x^2+3}+\frac{4-x^2}{x^2+8}+\frac{4-x^2}{x^2+10}=0\)

\(\Leftrightarrow\left(4-x^2\right)\left(\frac{1}{x^2+2}+\frac{1}{x^2+3}+\frac{1}{x^2+8}+\frac{1}{x^2+10}\right)=0\)

\(\Leftrightarrow4-x^2=0\)(do \(\frac{1}{x^2+2}+\frac{1}{x^2+3}+\frac{1}{x^2+8}+\frac{1}{x^2+10}>0,\forall x\))

\(\Leftrightarrow x^2=4\Leftrightarrow x=\pm2\)

\(KL...\)

2x(8x - 1)²(4x - 1) = 9

<=> 512x⁴ - 256x³ + 40x² - 2x = 9

<=> 512x⁴ - 256x³ + 40x² - 2x - 9 = 0

<=> (2x - 1)(4x + 1)(64x⁴ - 16x + 9) = 0

vì 64x⁴ - 16x + 9 khác 0 nên:

<=> 2x - 1 = 0 hoặc 4x + 1 = 0

<=> x = 1/2 hoặc x = -1/4

 (a+b)^3 - 3ab(a+b)
= (a^3 + 3a^2b + 3ab^2 + b^3) - (3a^2b + 3ab^2) 
= a^3 + 3a^2b + 3ab^2 + b^3 - 3a^2b - 3ab^2 
= a^3 + b^3 (đpcm) 

ta có

a³+b³=(a³+3a²b+3ab²+b³)-3a²b-3ab²

=(a+b)³-3ab(a+b) (ĐPCM)