a)\(\left(5x-1\right)^2-\left(5x-4\right)\left(5x+4\right)=7\)

\(\Leftrightarrow25x^2-10x+1-25x^2+16=7\)

\(\Leftrightarrow-10x=-10\)

\(\Leftrightarrow x=1\)

b) k hiểu đề

đề cũng là tìm x mà

a) (x - 7) + 5 = 16

=>   x - 7 = 11

=>  x  =  18

b) 4 : x - 34 = 6

=>  4 : x = 40

=>  x  =  \(\frac{1}{10}\)

c) 120 : ( x - 1 ) + 3 = 23

=>  120 : ( x - 1 ) = 20

=>  x - 1 = 6

=>  x  = 7

d) 170 - ( x-7) : 5 = 20

=>  ( x - 7 ) : 5 = 150

=>  x - 7  = 750

=>   x   =  757

e) x + 12 + 2 .x + 3 = 4.x + 3

=>  3x + 17  = 4x  + 3

=>  x  = 15 

\(a,x-7=11\)

\(x=18\)

\(b,x-36=\frac{2}{3}\)

\(x=\frac{110}{3}\)

\(c,120:\left(x-1\right)=20\)

\(x-1=6\)

\(x=7\)

\(d,\left(x-7\right):5=150\)

\(x-7=10\)

\(x=17\)

\(d,x+12+2.x+3-4.x-3=0\)

\(x.\left(1+2-4\right)+3-3+12=0\)

\(x.\left(-1\right)+12=0\)

\(x.\left(-1\right)=-12\)

\(x=12\)

a) \(5\times\left(3+7\times x\right)=400\)

\(3+7\times x=80\)

\(7\times x=77\)

\(x=11\)

b) \(x\times37+x\times63=1200\)

\(x\times\left(37+63\right)=1200\)

\(x\times100=1200\)

\(x=12\)

c) \(x\times6+12:3=40\)

\(x\times6+4=40\)

\(x\times6=36\)

\(x=6\)

d) \(4+6\times\left(x+1\right)=70\)

\(6\times\left(x+1\right)=66\)

\(x+1=11\)

\(x=10\)

e) \(163:x+34:x=10\)

\(\left(163+34\right):x=10\)

\(197:x=10\)

\(x=19,7\)

B1 :

a) \(\dfrac{2}{15}+3=\dfrac{2}{15}+\dfrac{15}{5}=\dfrac{17}{5}\)

b) \(2-\dfrac{7}{4}=\dfrac{8}{4}-\dfrac{7}{4}=\dfrac{1}{4}\)

c) \(\dfrac{17}{6}-\dfrac{11}{12}\times6=\dfrac{17}{6}-\dfrac{11}{2}=-\dfrac{8}{3}\)

d) \(\dfrac{14}{17}:7-\dfrac{15}{34}=\dfrac{2}{17}-\dfrac{5}{34}=\dfrac{4}{34}-\dfrac{5}{34}=-\dfrac{1}{34}\)

B3 :

\(x:23=146\)

\(x=146\times23\)

\(x=3358\)

b) \(x\times38=4066\)

\(x=4066:38\)

\(x=107\)

c) \(787+x\times67=2658\)

\(x\times67=2658-787\)

\(x\times67=1871\)

\(x=1871:67\)

\(x=\dfrac{1871}{67}\)

từng bài thôi 

a: ĐKXĐ của A là: \(\begin{cases}x+2<>0\\ x^2-4<>0\\ x^2+3x+2<>0\end{cases}\)

=>\(\begin{cases}x<>-2\\ x^2<>4\\ \left(x+1\right)\left(x+2\right)<>0\end{cases}\)

=>x∉{-2;2;-1}

ĐKXĐ cua B là \(x^3-1<>0\)

=>\(x^3<>1\)

=>x<>1

b: \(\frac{4x}{x+2}-\frac{x^3-8}{x^3+8}\cdot\frac{4x^2-8x+16}{x^2-4}\)

\(=\frac{4x}{x+2}-\frac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\cdot\frac{4\left(x^2-2x+4\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{4x}{x+2}-\frac{4\left(x^2+2x+4\right)}{\left(x+2\right)^2}=\frac{4x\left(x+2\right)-4x^2-8x-16}{\left(x+2\right)^2}\)

\(=\frac{4x^2+8x-4x^2-8x-16}{\left(x+2\right)^2}=-\frac{16}{\left(x+2\right)^2}\)

\(A=\left(\frac{4x}{x+2}-\frac{x^3-8}{x^3+8}\cdot\frac{4x^2-8x+16}{x^2-4}\right):\frac{16}{x+2}\cdot\frac{x^2+3x+2}{x^2+x+1}\)

\(=\frac{-16}{\left.\left(x+2\right)^2\right.}\cdot\frac{x+2}{16}\cdot\frac{\left(x+1\right)\left(x+2\right)}{x^2+x+1}=\frac{-\left(x+1\right)}{x^2+x+1}\)

\(B=\frac{x^2+x-2}{x^3-1}\)

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

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

b: Đặt P=A+B

\(=\frac{x+2-x-1}{x^2+x+1}=\frac{1}{x^2+x+1}\)

\(=\frac{1}{x^2+x+\frac14+\frac34}=\frac{1}{\left(x+\frac12\right)^2+\frac34}\le1:\frac34=\frac43\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi x+1/2=0

=>x=-1/2

\(-2.x-\left(x-170\right)=34-\left(x+25\right)\)

       \(-2x-x+170=34+x-25\)

               \(-3x+170=9+x\)

                     \(-3x-x=-9-170\)

                             \(-4x=-179\)

                                    \(x=\frac{179}{4}\)