Giup mk vs mk sap phai nop roi
\(\left(2\times x-y\right)^5+\left(2\times y-8\right)^{2014}=0\).Tim x,y
\(\left(\frac{1}{2}.x-5\right)^{29}+\left(y^2-\frac{1}{4}\right)^{10}< hoac=0\)0
help me giup mk voi cau xin cac bn
mai mk nop roi
\(\left(\frac{1}{2}x-5\right)^{29}\)ko làm đc
Phải mũ chẵn mới ra
\(\left(2\times x-y\right)^5+\left(2\times y-8\right)^{2014}=0\)
\(\dfrac{1}{3}x-\dfrac{2}{5}\left(x+1\right)=0\)
\(\dfrac{x+2}{0,5}=\dfrac{2x+1}{2}\)
Tim x
giai giup nhe, mai phai nop roi :(
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
\(\left(2\times X-y\right)^5+\left(2\times y-8\right)=0\\ \)
Tìm x,y biết
\(\left(x-3\right)^2+\left(y+2\right)^2=0\)
\(2\times x+2^{x+3}=136\)
\(\left(x-12+y\right)^{200}+\left(x-4-y\right)^{200}=0\)
\(\left(2\times x-5\right)^{2000}+\left(3\times y+4\right)^{2002}\le0\)
\(\left(x-3\right)^2+\left(y+2\right)^2=0\)
\(\left\{{}\begin{matrix}\left(x-3\right)^2\ge0\forall x\\\left(y+2\right)^2\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow\left(x-3\right)^2+\left(y+2\right)^2\ge0\)
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 \(\left\{{}\begin{matrix}\left(x-12+y\right)^{200}\ge0\forall x,y\\\left(x-4-y\right)^{200}\ge0\forall x,y\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x-12+y\right)^{200}=0\\\left(x-4-y\right)^{200}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-12+y=0\\x-4-y=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=12\\x-y=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=8\\y=4\end{matrix}\right.\)
\(\left(^{x^2}\times y\right)^{^5}\times\left(x^2\times y^2\right)^7\times\left(x\times y^2\right)^6\times x^3\)
\(\left(x^2.y\right)^5.\left(x^2.y^2\right)^7.\left(x.y^2\right)^6.x^3\)
\(=x^{10}.y^5.x^{14}.y^{14}.x^6.y^{12}.x^3\)
\(=x^{33}.y^{31}\)
ban nao onl thay bai nay giai gap giup mk nha!
Tim x,y, bt:
\(\left(x-5\right)^{88}+\left(x+y+3\right)^{496}>=\left(lonhonhoacbang\right)0\)
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!
1 tim x \(2014.\left|x-12\right|+\left(x-12\right)^2=2013.\left|12-x\right|\)\(x\)|
2 chung minh \(8^7-2^{18}⋮14\)
3 tim x,y,z biet 4x=7y=3z va x+y+z=61
4 tim a,b,c biet \(\frac{1}{2}a=\frac{2}{3}b=\frac{3}{4}c\)vs \(a-b=15\)
giup mk nha moi nguoi,lm dc cang nhiu cang tot
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:
\(\frac{1}{2}a=\frac{2}{3}b=\frac{3}{4}c\Rightarrow\frac{a}{2}=\frac{2b}{3}=\frac{3c}{4}\Rightarrow\frac{a}{2.6}=\frac{2b}{3.6}=\frac{3c}{4.6}\Rightarrow\frac{a}{12}=\frac{b}{9}=\frac{c}{8}=\frac{a-b}{12-9}=\frac{15}{3}=5\)
\(\Rightarrow a=5.12=60,b=9.5=45,c=8.5=40\)
Tìm x,y biết:
\(\left|x+\frac{11}{7}\right|+\left|x+\frac{2}{7}\right|+\left|x+\frac{4}{7}\right|=4x\)
\(\left(x-2\right)^{2014}+\left(y-1\right)^{2016}+\left(x-y-2\right)=0\)
mk đang cần gấp giúp mk nha các bn