các bạn đi học chưa hả?

Chưa bạn

bạn ở đâu vậy?

\(A=\frac{1}{x+5}+\frac{2}{x-5}-\frac{2x+10}{\left(x+5\right)\left(x-5\right)}\)   ĐK đề bài

\(=\frac{x-5+2\left(x+5\right)-2x-10}{\left(x+5\right)\left(x-5\right)}=\frac{-\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}=-\frac{1}{x-5}\)

b/ có A=-3 => \(-\frac{1}{x-5}=-3 \Rightarrow x-5=\frac{1}{3}\Rightarrow x=\frac{16}{3}\)

có \(9x^2-42x+49=\left(3x-7\right)^2=\left(\frac{3.16}{3}-7\right)^2=81\)

\(4x^2-28x+49=\left(2x\right)^2-2\cdot2x\cdot7+7^2=\left(2x-7\right)^2\)

thay x=4 vào ta được \(\left(2\cdot4-7\right)^2=\left(8-7\right)^2=1^2=1\)

vậy \(4x^2-28x+49=1\)khi x=4

\(9x^2+42x+49=\left(3x\right)^2+2\cdot3x\cdot7+7^2=\left(3x+7\right)^2\)

thay x=1 và ta được \(\left(3\cdot1+7\right)^2=10^2=100\)

vậy \(9x^2+42x+49=100\)đạt được khi x=1

\(25x^2-2xy+\frac{1}{25y^2}=\left(5x\right)^2-2\cdot5x\cdot\frac{1}{5y}+\left(\frac{1}{5y}\right)^2=\left(5x-\frac{1}{5y}\right)^2\)

thay x=\(\frac{-1}{5}\)và y=-5 vào ta được \(\left[5\cdot\left(\frac{-1}{5}\right)-\frac{1}{5\cdot\left(-5\right)}\right]^2=\left(1-\frac{1}{-25}\right)^2=\left(\frac{26}{25}\right)^2=...\)

vậy \(25x^2-2xy+\frac{1}{25y^2}=\left(\frac{26}{25}\right)^2\)khi x=\(\frac{-1}{5}\)và y=-5

4x² - 28x + 49 = ( 2x )² - 2.2x.7 + 7² = ( 2x - 7 )²

Thế x = 4 ta được : ( 2 . 4 - 7 )² = 1² = 1

9x² + 42x + 49 = ( 3x )² + 2.3x.7 + 7² = ( 3x + 7 )²

Thế x = 1 ta được : ( 3.1 + 7 )² = 10² = 100

25x² - 2xy + 1/25y² = ( 5x )² - 2.5x.1/5y + ( 1/5y )² = ( 5x - 1/5y )²

Thế x = -1/5 , y = -5 ta được : \(\left[5\cdot\left(-\frac{1}{5}\right)-\frac{1}{5}\cdot\left(-5\right)\right]^2=\left[-1+1\right]^2=0\)

a) \(A=\dfrac{1}{x+5}+\dfrac{2}{x-5}-\dfrac{2x+10}{\left(x+5\right)\left(x-5\right)}\)

\(A=\dfrac{x-5+2x+10-2x-10}{\left(x+5\right)\left(x-5\right)}=\dfrac{x-5}{\left(x+5\right)\left(x-5\right)}=\dfrac{1}{x+5}\)

b) \(A=-3\Rightarrow\dfrac{1}{x+5}=-3\)

\(\Leftrightarrow x+5=-\dfrac{1}{3}\Leftrightarrow x=-\dfrac{1}{3}-5=\dfrac{-16}{3}\)

\(9x^2-42x+49=\left(3x-7\right)^2=\left(3.\dfrac{-16}{3}-7\right)^2=\left(-23\right)^2=529\) \(\left(x=\dfrac{-16}{3}\right)\)

Bài 1:

a)  ĐKXĐ:  \(x\ne\pm5\)

\(A=\frac{1}{x+5}+\frac{2}{x-5}-\frac{2x+10}{\left(x+5\right)\left(x-5\right)}\)

\(=\frac{x-5}{\left(x+5\right)\left(x-5\right)}+\frac{2\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\frac{2x+10}{\left(x-5\right)\left(x+5\right)}\)

\(=\frac{x-5+\left(2x+10\right)-\left(2x+10\right)}{\left(x-5\right)\left(x+5\right)}\)

\(=\frac{x-5}{\left(x-5\right)\left(x+5\right)}=\frac{1}{x+5}\)

b)  \(B=9x^2-42x+49=\left(3x-7\right)^2\)

Tại  \(x=-3\)thì:   \(B=\left[3.\left(-3\right)-7\right]^2=256\)

Bài 2:

a)  ĐKXĐ:  \(x\ne\pm3\)

\(A=\frac{3}{x+3}+\frac{1}{x-3}-\frac{18}{9-x^2}\)

\(=\frac{3\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{x+3}{\left(x-3\right)\left(x+3\right)}+\frac{18}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{3x-9+x+3+18}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{4x+12}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{4\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{4}{x-3}\)

b)  \(A=4\)\(\Rightarrow\)\(\frac{4}{x-3}=4\)

\(\Rightarrow\)\(4\left(x-3\right)=4\)\(\Leftrightarrow\)\(x-3=1\)\(\Leftrightarrow\)\(x=4\)   (t/m ĐKXĐ)

Vậy....

\(9x^2+42x+49=\left(3x+7\right)^2\)

Thay x=1 ta có 

\(\left(3.1+7\right)^2=10^2=100\)

\(\frac{1}{4}a^2+2ab^2+4b^4=\left(\frac{1}{2}a+2b^2\right)^2\)

Thay a=2;b=-1 ta có 

\(\left(\frac{1}{2}.2+2\left(-1\right)^2\right)^2=\left(1+2\right)^2=3^2=9\)

\(\(9x^2+42x+49\)\)tại x = 1

Ta có:\(\(9x^2+42x+49=\left(3x\right)^2+2.3x.7+7^2=\left(3x+7\right)^2\)\)

Thay x = 1 vào \(\(\left(3x+7\right)^2\)\)ta được:

\(\(\left(3.1+7\right)^2=10^2=100\)\)

\(\(\frac{1}{4}a^2+2ab^2+4b^4\)\)tại a = 2 ; b = -1

Ta có: \(\frac{1}{4}a^2+2ab^2+4b^4=\left(\frac{1}{2}a\right)^2+2.\frac{1}{2}a.2b^2+\left(2b\right)^2=\left(\frac{1}{2}a+2b^2\right)^2\)

Thay a = 2 ; b = -1 vào\(\left(\frac{1}{2}a+2b^2\right)^2\)ta được:

\(\(\left(\frac{1}{2}.2+2.\left(-1\right)^2\right)^2=\left(3\right)^2=9\)\)

1. ĐKXĐ: \(x\ne\pm1\)

2. \(A=\left(\dfrac{x+1}{x-1}-\dfrac{x+3}{x+1}\right)\cdot\dfrac{x+1}{2}\)

\(=\dfrac{\left(x+1\right)^2-\left(x-3\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{2}\)

\(=\dfrac{x^2+2x+1-x^2+4x-3}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{2}\)

\(=\dfrac{6x-2}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{2}\)

\(=\dfrac{2\left(x-3\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{x-3}{x-1}\)

3. Tại x = 5, A có giá trị là:

\(\dfrac{5-3}{5-1}=\dfrac{1}{2}\)

4. \(A=\dfrac{x-3}{x-1}\) \(=\dfrac{x-1-3}{x-1}=1-\dfrac{3}{x-1}\)

Để A nguyên => \(3⋮\left(x-1\right)\) hay \(\left(x-1\right)\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)

\(\Rightarrow\left\{{}\begin{matrix}x-1=1\\x-1=-1\\x-1=3\\x-1=-3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\left(tmđk\right)\\x=0\left(tmđk\right)\\x=4\left(tmđk\right)\\x=-2\left(tmđk\right)\end{matrix}\right.\)

Vậy: A nguyên khi \(x=\left\{2;0;4;-2\right\}\)