mng giúp với

\(\Leftrightarrow7x\left(x+5\right)+\left(x-5\right)\left(x+5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(7x+x+5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(8x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-\dfrac{5}{8}\end{matrix}\right.\)

\(\Leftrightarrow x^2+x-5x-5-x^2-6x-9-6=0\\ \Leftrightarrow-10x-20=0\\ \Leftrightarrow x=-2\)

a) x=60

b) x=107

f) x=2

.....

CHO MÌNH BIẾT CẢ CÁCH LÀM NHA

a) 315 + ( 146 - x ) = 401

<=> 146 - x = 401 - 315

<=> 146 - x = 86

<=> x = 60

Vậy x  = 60

         \(x^2-5x-4\left(x-5\right)=0\)

\(\Leftrightarrow\)\(x\left(x-5\right)-4\left(x-5\right)=0\)

\(\Leftrightarrow\)\(\left(x-5\right)\left(x-4\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x-5=0\\x-4=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=5\\x=4\end{cases}}\)

Vậy....

\(2x\left(x+6\right)=7x+42\)

\(\Leftrightarrow\)\(2x\left(x+6\right)-7x-42=0\)

\(\Leftrightarrow\)\(2x\left(x+6\right)-7\left(x+6\right)=0\)

\(\Leftrightarrow\)\(\left(x+6\right)\left(2x-7\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x+6=0\\2x-7=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-6\\x=\frac{7}{2}\end{cases}}\)

Vậy......

\(x^3-5x^2+x-5=0\)

\(\Leftrightarrow\)\(x^2\left(x-5\right)+\left(x-5\right)=0\)

\(\Leftrightarrow\)\(\left(x-5\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow\)\(x-5=0\)

\(\Leftrightarrow\)\(x=5\)

\(x^4-2x^3+10x^2-20x=0\)

\(\Leftrightarrow\)\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)

\(\Leftrightarrow\)\(x\left(x-2\right)\left(x^2+10\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

Vậy...

Bài 2: 

a: \(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)

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

a) \(7x\left(x+1\right)-3\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(7x-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\7x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{3}{7}\end{matrix}\right.\)

b) 3(x + 8) - x² - 8x = 0

=> 3(x + 8) - (x² + 8x) = 0

=> 3(x + 8) - x(x + 8) = 0

=> (x + 8)(3 - x) = 0 => \(\left[{}\begin{matrix}x+8=0\\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-8\\x=3\end{matrix}\right.\)

c) \(x^2-10x=-25\Rightarrow x^2-10x+25=0\Rightarrow\left(x-5\right)^2=0\Rightarrow x=5\)

d) Giống câu c

⇒[x+1=07x+3=0⇒⎡⎣x=−1x=−37⇒[x+1=07x+3=0⇒[x=−1x=−37

b) 3(x + 8) - x² - 8x = 0

=> 3(x + 8) - (x² + 8x) = 0

=> 3(x + 8) - x(x + 8) = 0

=> (x + 8)(3 - x) = 0 => [x+8=03−x=0⇒[x=−8x=3[x+8=03−x=0⇒[x=−8x=3

c) ${}^{}-10x=-25\Rightarrow {}^{}-10x+$

a. \(x^2-25-3.\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x+5\right)-3.\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x+5-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

b. \(\left(3x+1\right)^2=\left(2x-5\right)\\ \Leftrightarrow9x^2+6x+1=2x-5\\ \Leftrightarrow9x^2+6x-2x=-5-1\\ \Leftrightarrow9x^2+4x=-6\\ \Leftrightarrow x\left(9x+4\right)=-6\\ \Leftrightarrow\left[{}\begin{matrix}x=-6\\9x+4=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=-\dfrac{10}{9}\end{matrix}\right.\)

c. \(2x^2-7x+6=0\\ \Leftrightarrow2x^2-7x=-6\\ \Leftrightarrow x\left(2x-7\right)=-6\\ \Leftrightarrow\left[{}\begin{matrix}x=-6\\x=\dfrac{1}{2}\end{matrix}\right.\)

a, \(\left(x-5\right)\left(x+5\right)-3\left(x-5\right)=0\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\Leftrightarrow x=-2;x=5\)

b, bạn ktra lại đề, thường thường ngta hay cho 2 vế cùng bình phương 

c, \(2x^2-7x+6=0\Leftrightarrow\left(2x-3\right)\left(x-2\right)=0\Leftrightarrow x=\dfrac{3}{2};x=2\)

\(\left(3x+1\right)^2=\left(2x-5\right)^2\)

\(\left[{}\begin{matrix}3x+1=2x-5\\3x+1=5-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\5x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=\dfrac{4}{5}\end{matrix}\right.\)