P= 7x²-3xy³+4
P= 7.(-2)²-3.(-2).(-1/2)³+4
=7.4-3.(-2).(-1/8)+4
=28-(-6).(-1/8)+4
=28-3/4+4
=109/4+4
=125/4
kb đúng hay sai.

Đề bài yêu cầu j vậy?

mũ 2 tui ko bt

co ai giai dc ko cau 60 % .....

$\textbf{Ta có:}$

$A=\dfrac{-5^2-5\cdot3^2}{5^3+5^2\cdot3^2}$

$=\dfrac{-25-45}{125+225}$

$=\dfrac{-70}{350}$

$=-\dfrac15.$

Và

$B=\dfrac{2^{12}\cdot3^{10}+6^9\cdot120}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}$

$=\dfrac{2^{12}3^{10}+2^{12}3^{10}\cdot5}{2^{11}3^{11}(2\cdot3-1)}$

$=\dfrac{2^{12}3^{10}(1+5)}{2^{11}3^{11}\cdot5}$

$=\dfrac{2^{12}3^{10}\cdot6}{2^{11}3^{11}\cdot5}$

$=\dfrac{2^2}{5}$

$=\dfrac45.$

=> $M=B-A$$=\dfrac45-\left(-\dfrac15\right)$

$=\dfrac55$$=1.$

\(\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{z}{2}\)

\(\Rightarrow\dfrac{x^3}{125}=\dfrac{y^3}{64}=\dfrac{z^3}{8}=\dfrac{x^3-y^3+z^3}{125-64+8}=\dfrac{69}{69}=1\)

\(\Rightarrow\left\{{}\begin{matrix}x=\sqrt[3]{125}=5\\y=\sqrt[3]{64}=4\\z=\sqrt[3]{8}=2\end{matrix}\right.\)

x bằng bao nhiêu cậu ơi ?

x=\(\frac{y}{2}\)=\(\frac{z}{3}\)

\(P\left(x\right)=x^2+5x^4-3x^3+x^2+4x^4+3x^3-x+5\)

\(=\left(5x^4+4x^4\right)+\left(-3x^3+3x^3\right)+\left(x^2+x^2\right)-x+5\)

\(=9x^4+2x^2-x+5\)

\(Q\left(x\right)=x-5x_{}^3-x^2-x^4+4x^3-x^2+3x-1\)

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

\(=-6x^4+4x^3-2x^2+4x-1\)

P(x)+Q(x)

\(=9x^4+2x^2-x+5-6x^4+4x^3-2x^2+4x-1\)

\(=3x^4+4x^3+3x+4\)

Đặt P(x)+Q(x)=0

=>\(3x^4+4x^3+3x+4=0\)

=>\(x^3\left(3x+4\right)+\left(3x+4\right)=0\)

=>\(\left(3x+4\right)\left(x^3+1\right)=0\)

=>\(\left[\begin{array}{l}3x+4=0\\ x^3+1=0\end{array}\right.\Longrightarrow\left[\begin{array}{l}x=-\frac43\\ x=-1\end{array}\right.\)

\(A=\frac{1}{2}-\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3-\left(\frac{1}{2}\right)^4+...-\left(\frac{1}{2}\right)^{20}\)

\(2A=1-\frac{1}{2}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^3+...-\left(\frac{1}{2}\right)^{19}\)

\(2A-A=\)\(\left(1-\frac{1}{2}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^3+...-\left(\frac{1}{2}\right)^{19}\right)-\)\(\left(\frac{1}{2}-\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3-\left(\frac{1}{2}\right)^4+...-\left(\frac{1}{2}\right)^{20}\right)\)

\(A=1-\left(\frac{1}{2}\right)^{20}\)