\(\left(\frac{1}{2}x-5\right)^{29}\)ko làm đc

Phải mũ chẵn mới ra

la sao bn

là bn có viết sai đề không ấy

a: =>1/3x-2/5x-2/5=0

=>-1/15x=2/5

hay x=-6

b: =>2(x+2)=0,5(2x+1)

=>2x+4=x+0,5

=>x=-3,5

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

\(\left\{{}\begin{matrix}\left(x-3\right)^2\ge0\forall x\\\left(y+2\right)^2\ge0\forall y\end{matrix}\right.\)

\(\Rightarrow\left(x-3\right)^2+\left(y+2\right)^2\ge0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left(x-3\right)^2=0\Rightarrow x-3=0\Rightarrow x=3\\\left(y+2\right)^2=0\Rightarrow y+2=0\Rightarrow y=-2\end{matrix}\right.\)

đề sai câu b các câu sau áp dụng tương tự

c/ Vì: \(\left(x-12+y\right)^{200}+\left(x-4-x\right)^{200}=0\)

mà \(\left\{{}\begin{matrix}\left(x-12+y\right)^{200}\ge0\forall x,y\\\left(x-4-y\right)^{200}\ge0\forall x,y\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x-12+y\right)^{200}=0\\\left(x-4-y\right)^{200}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-12+y=0\\x-4-y=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x+y=12\\x-y=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=8\\y=4\end{matrix}\right.\)

\(\left(x^2.y\right)^5.\left(x^2.y^2\right)^7.\left(x.y^2\right)^6.x^3\)

\(=x^{10}.y^5.x^{14}.y^{14}.x^6.y^{12}.x^3\)

\(=x^{33}.y^{31}\)

Ta thấy: \(\left(x-5\right)^{88}\ge0\) 

\(\left(x+y+3\right)^{496}\ge0\) 

\(\Rightarrow\left(x-5\right)+\left(x+y+3\right)^{496}\ge\) ( Đó là điều đương nhiên )

Vậy: \(x;y\in R\)

\(\left(x-5\right)^{88}+\left(x+y+z\right)^{496}\ge0\)0

Dấu "=" xảy ra kih và chỉ khi \(\hept{\begin{cases}\left(x-5\right)^{88}\\\left(x+y+3\right)^{496}\end{cases}\Leftrightarrow\hept{\begin{cases}x=5\\5+y+3=0\end{cases}}}\)\(\Leftrightarrow\hept{\begin{cases}x=5\\y=-8\end{cases}}\)

Mk chi p trg hop bang 0 thoi

Neu: \(\left(x-5\right)^{88}+\left(x+y+3\right)^{496}\)=0

Thi: \(\left(x-5\right)^{88}\)=\(0\)

ma x mu may cung bang 0

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

=> \(x+y+3=0\)

Ma \(x=5\)

nen \(x+y+3=5+y+3=0\)

=> \(y=-8\)

Vay \(x=5\)\(,y=-8\)

Trg hop lon hon 0 thi chac la hk co!
 

câu 1: Câu hỏi của Vương Ái Như - Toán lớp 7 - Học toán với OnlineMath

câu 2:

Ta có: \(8^7-2^{18}=2^{21}-2^{18}=2^{17}.\left(2^4-2\right)=2^{17}.14⋮14\)

câu 3:

\(4x=7y=3x\Rightarrow\frac{4x}{84}=\frac{7y}{84}=\frac{3z}{84}\Rightarrow\frac{x}{21}=\frac{y}{12}=\frac{z}{28}=\frac{x+y+z}{21+12+28}=\frac{61}{61}=1\)

\(\Rightarrow x=21,y=12,z=28\)

câu 4:

\(\frac{1}{2}a=\frac{2}{3}b=\frac{3}{4}c\Rightarrow\frac{a}{2}=\frac{2b}{3}=\frac{3c}{4}\Rightarrow\frac{a}{2.6}=\frac{2b}{3.6}=\frac{3c}{4.6}\Rightarrow\frac{a}{12}=\frac{b}{9}=\frac{c}{8}=\frac{a-b}{12-9}=\frac{15}{3}=5\)

\(\Rightarrow a=5.12=60,b=9.5=45,c=8.5=40\)