tớ sẽ cho kết quả luôn:

A)258,888

B)356,216

nhớ k cho tớ nhé

Tính giá trị của biểu thức:

a) Tại \(y=3\) ta có 

\(\left(3\right)^3-21.3+3\) = \(27-63+3\) = -33

b) Tại \(x=5\) và \(y=2\) ta có 

\(5.\left(2\right)^2-12.5.2+(5)^2\) = \(5.4-120+25\) = \(20-120+25=-75\)

\(a^2-2a+6b+b^2=-10\\ \Leftrightarrow a^2-2a+1+b^2+6b+9=0\\ \Leftrightarrow\left(a-1\right)^2+\left(b+3\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}a=1\\b=-3\end{matrix}\right.\)

Vậy \(\left(a;b\right)=\left(1;-3\right)\)

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\\ \Leftrightarrow xy+yz+zx=0\\ \Rightarrow\left\{{}\begin{matrix}xy+yz=-zx\\xy+zx=-yz\\yz+zx=-xy\end{matrix}\right.\)

Ta có: 

\(A=\dfrac{xz+yz}{z^2}+\dfrac{xy+yz}{y^2}+\dfrac{xy+xz}{x^2}\\ =\dfrac{-xy}{z^2}+\dfrac{-xz}{y^2}+\dfrac{-yz}{x^2}\\ =-xyz\cdot\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)\\ =-xyz\cdot\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}-\dfrac{2}{xy}-\dfrac{2}{yz}-\dfrac{2}{xz}\right)\\ =0\)

\(A=\left(X+3\right)^3\)

Thay x = - 3 vào, ta có:

\(A=\left(-3+3\right)^3=0\)

B=50x49+53x50

B=50x(49+53)

B=50x102

vậy A<B vì 50x101<50x102

Bg

Ta có: A = 101.50 và B = 50.49 + 53.50

Xét B = 50.49 + 53.50:

=> B = 50.(49 + 53)

=> B = 50.102

Vì 101.50 < 50.102 nên A < B

Vậy A < B

a: ĐKXĐ: x<>2; x<>-2

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

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

\(=\frac{x-2\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\frac{x+2}{2}=\frac{2x-2-2x-4}{2\left(x-2\right)}=\frac{-6}{2\left(x-2\right)}\)

\(=\frac{-3}{x-2}\)

b: Để A nguyên thì -3⋮x-2

=>x-2∈{1;-1;3;-3}

=>x∈{3;1;5;-1}

a: ĐKXĐ: x<>0; x<>1; x<>-1

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

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

\(=\frac{x^2+2x+1-x^2+2x-1}{x+1}\cdot\frac{5}{2x}=\frac{4x}{x+1}\cdot\frac{5}{2x+1}=2\cdot\frac{5}{x+1}=\frac{10}{x+1}\)

c: Vì x=-1 không thỏa mãn ĐKXĐ

nên khi x=-1 thì A không có giá trị

THay x=3 vào A, ta được:

\(A=\frac{10}{3+1}=\frac{10}{4}=\frac52\)