Tl :

Cho anh hỏi Cạnh bên dài 8 cm

thì cạnh bên là cạnh nào

k bt đc

vì tam giác  cân nên là cạnh bên bắng nhau thế thì cạnh bên nào bằng tám mà chả dc

a) \(A=5x\left(4x^2-2x+1\right)-2x\left(10x^2-5x-2\right)\)

\(A=20x^3-10x^2+5x-20x^3+10x^2+4x\)

\(A=9x\)

Thay x = 15 vào, ta có: 

\(A=9.15=135\)

b) \(B=5x\left(x-4y\right)-4y\left(y-5x\right)\)

\(B=5x^2-20xy-4y^2+20xy\)

\(B=5x^2-4y\)

Thay \(x=-\frac{1}{5};y=-\frac{1}{2}\) vào, ta có: 

\(B=5.\left(-\frac{1}{5}\right)^2-4.\left(-\frac{1}{2}\right)=\frac{11}{5}\)

c) \(C=6xy\left(xy-y^2\right)-8x^2\left(x-y^2\right)-5y^2\left(x^2-xy\right)\)

\(C=6x^2y^2-6xy^3-8x^3+8x^2y^2-5x^2y^2+5xy^3\)

\(C=9x^2y^2-xy^3-8x^3\)

Thay \(x=\frac{1}{2};y=2\) vào, ta có:

\(C=9.\left(\frac{1}{2}\right)^2.2^2-\frac{1}{2}.2^3-8.\left(\frac{1}{2}\right)^3=4\)

d) \(D=\left(3x+5\right)\left(2x-1\right)+\left(4x-1\right)\left(3x+2\right)\)

\(D=6x^2-3x+10x-5+12x^2+8x-3x-2\)

\(D=18x^2+12x-7\)

Ta có: \(\left|2\right|=\orbr{\begin{cases}x=-2\\x=2\end{cases}}\)

+) Với x = -2

\(D=18.\left(-2\right)^2+12.\left(-2\right)-7=41\)

+) Với x = 2

\(D=18.2^2+12.2-7=89\)

\(\Leftrightarrow\left(x-3\right)^3-\left(x-3\right)^3+9\left(x^2+1\right)=63\)

\(\Leftrightarrow9\left(x^2+1\right)=63\)

\(\Leftrightarrow x^2+1=7\)

\(\Leftrightarrow x^2=6\)

\(\Leftrightarrow x=\pm\sqrt{6}\)

ta có: ( x –3 )³ – ( x – 3) . ( x² + 3x + 9) + 9 (x² + 1 ) = 63

\(\Leftrightarrow x^3-3.x^2.3+3.x.3^2-3^3+3.\left(x^2+3x+9\right)-x\left(x^2+3x+9\right)+9x^2+\)\(9=63\)

\(\Leftrightarrow\left(x^3-x^3\right)-\left(9x^2-9x^2\right)+27x-\left(27-27\right)-\left(9x-9x\right)-\left(3x^2-3x^2\right)\)\(+9=63\)

\(\Leftrightarrow27x+9=63\)

\(\Leftrightarrow3x+1=7\)

\(\Leftrightarrow3x=6\)

\(\Leftrightarrow x=2\)

vậy: x=2

a) \(ĐKXĐ:x\ne\pm1\)

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

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

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

b) Để A có giá trị nguyên

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

\(\Leftrightarrow x^2-x⋮x+1\)

\(\Leftrightarrow x^2-x-2+2⋮x+1\)

\(\Leftrightarrow\left(x+1\right)\left(x-2\right)+2⋮x+1\)

\(\Leftrightarrow2⋮x+1\)

\(\Leftrightarrow x+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

\(\Leftrightarrow x\in\left\{-2;0;-3;1\right\}\)

Ta sẽ loại các giá trị ktm

\(\Leftrightarrow x\in\left\{-2;0;-3\right\}\)

Vậy để \(A\inℤ\Leftrightarrow x\in\left\{-2;0;-3\right\}\)

Từ giả thiết suy ra \(\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{c+a}{ca}\)(vì a,b,c khác 0)

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

\(\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\Rightarrow a=b=c\)

\(\Rightarrow M=1\)