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a) Ta có: \(\left|2x-5\right|\ge0\forall x\)
\(\left|3y+1\right|\ge0\forall y\)
Do đó: \(\left|2x-5\right|+\left|3y+1\right|\ge0\forall x,y\)
mà \(\left|2x-5\right|+\left|3y+1\right|=0\)
nên \(\left\{{}\begin{matrix}\left|2x-5\right|=0\\\left|3y+1\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-5=0\\3y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=5\\3y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{5}{2}\\y=\frac{-1}{3}\end{matrix}\right.\)
Vậy: \(x=\frac{5}{2}\) và \(y=\frac{-1}{3}\)
b) Ta có: \(\left|3x-4\right|\ge0\forall x\)
\(\left|3y-5\right|\ge0\forall y\)
Do đó: \(\left|3x-4\right|+\left|3y-5\right|\ge0\forall x,y\)
mà \(\left|3x-4\right|+\left|3y-5\right|=0\)
nên \(\left\{{}\begin{matrix}\left|3x-4\right|=0\\\left|3y-5\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-4=0\\3y-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=4\\3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{4}{3}\\y=\frac{5}{3}\end{matrix}\right.\)
Vậy: \(x=\frac{4}{3}\) và \(y=\frac{5}{3}\)
c) Ta có: |16-|x||≥0∀x
\(\left|5y-2\right|\ge0\forall y\)
Do đó: |16-|x||+|5y-2|≥0∀x,y
mà |16-|x||+|5y-2|=0
nên \(\left\{{}\begin{matrix}\text{|16-|x||}=0\\\left|5y-2\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}16-\left|x\right|=0\\5y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left|x\right|=16\\5y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{16;-16\right\}\\y=\frac{2}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{16;-16\right\}\) và \(y=\frac{2}{5}\)
có |2x-5| luôn \(\ge0\forall x\in Q\)
cũng có \(\left|3y+1\right|\ge0\forall y\in Q\)
=> \(\left|2x-5\right|+\left|3y-1\right|\ge0\forall x;y\in Q\)
=>\(\hept{\begin{cases}2x-5=0\\3y-1=0\end{cases}}\)<=> \(\hept{\begin{cases}2x=5\\3y=1\end{cases}}\)<=> \(\hept{\begin{cases}x=\frac{2}{5}\\y=\frac{1}{3}\end{cases}}\)
vậy \(x=\frac{2}{5};y=\frac{1}{3}\)
em nhớ là phải dùng ngoặc nhọn như trên nhé! Nếu không sẽ sai đấy!
3 câu còn lại cũng tương tự
1) a) \(\left|7x-5y\right|+\left|2z-3y\right|+\left|xy+yz+xz-2000\right|\ge0\)
Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}7x=5y\\2z=3y\\xy+yz+xz=2000\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{7}y\\z=\dfrac{3}{2}y\\xy+yz+xz=2000\end{matrix}\right.\)
Ta có: \(xy+yz+xz=2000\)
\(\Rightarrow\dfrac{5}{7}y^2+\dfrac{3}{2}y^2+\dfrac{15}{14}y^2=2000\)
\(\Rightarrow y^2\left(\dfrac{5}{7}+\dfrac{3}{2}+\dfrac{15}{14}\right)=2000\Leftrightarrow\dfrac{23}{7}y^2=2000\)
Tìm \(y\) và suy ra \(x;z\) là được,Bài này nghiệm khá xấu
b) \(\left|3x-7\right|+\left|3x+2\right|+8=\left|7-3x\right|+\left|3x+2\right|+8\ge\left|7-3x+3x+2\right|+8\ge9+8=17\)Dấu "=" xảy ra khi: \(-\dfrac{3}{2}\le x\le\dfrac{7}{3}\)
2) a)Ta có: \(\left\{{}\begin{matrix}\left|x-5\right|+\left|1-x\right|\ge\left|x-5+1-x\right|=4\\\dfrac{12}{\left|y+1\right|+3}\le\dfrac{12}{3}=4\end{matrix}\right.\)
Mà theo đề bài: \(\left|x-5\right|+\left|1-x\right|=\dfrac{12}{\left|y+1\right|+3}\)
\(\Rightarrow\left|x-5\right|+\left|1-x\right|=\dfrac{12}{\left|y+1\right|+3}=4\)
\(\Rightarrow\left\{{}\begin{matrix}1\le x\le5\\y=-1\end{matrix}\right.\)
b) Ta có: \(\left\{{}\begin{matrix}\left|y+3\right|+5\ge5\\\dfrac{10}{\left(2x-6\right)^2+2}\le\dfrac{10}{2}=5\end{matrix}\right.\)
Mà theo đề bài: \(\left|y+3\right|+5=\dfrac{10}{\left(2x-6\right)^2+2}\)
\(\Rightarrow\left|y+3\right|+5=\dfrac{10}{\left(2x-6\right)^2+2}=5\)
\(\Rightarrow\left\{{}\begin{matrix}y=-3\\x=3\end{matrix}\right.\)
c) Ta có: \(\left\{{}\begin{matrix}\left|x-1\right|+\left|3-x\right|\ge\left|x-1+3-x\right|=2\\\dfrac{6}{\left|y+3\right|+3}\le\dfrac{6}{3}=2\end{matrix}\right.\)
Mà theo đề bài: \(\left|x-1\right|+\left|3-x\right|=\dfrac{6}{\left|y+3\right|+3}\)
\(\Rightarrow\left|x-1\right|+\left|3-x\right|=\dfrac{6}{\left|y+3\right|+3}=2\)
\(\Rightarrow\left\{{}\begin{matrix}1\le x\le3\\y=-3\end{matrix}\right.\)
Ta có \(\hept{\begin{cases}\left(3x-5\right)^{2008}\ge0\\\left(y^2-1\right)^{2010}\ge0\\\left(x-z\right)^{2012}\ge0\end{cases}}\)mà \(\left(3x-5\right)^{2008}+\left(y^2-1\right)^{2010}+\left(x-z\right)^{2012}=0\)
\(\Rightarrow\hept{\begin{cases}\left(3x-5\right)^{2008}=0\\\left(y^2-1\right)^{2010}=0\\\left(x-z\right)^{2012}=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}3x-5=0\\y^2-1=0\\x-z=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{5}{3}\\y=1;-1\\z=x=\frac{5}{3}\end{cases}}\)
Bài 1 :\(a,=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{100^2}{99.101}\)
\(=\frac{2.3.4...100}{1.2.3...99}.\frac{2.3.4...100}{3.4...101}\)
\(=100.\frac{2}{101}=\frac{200}{101}\)
a) \(x^2=x^5\Rightarrow x^2=x^2.x^3\Rightarrow1=x^3\Rightarrow1^3=x^3\Rightarrow x=1\)
b)\(y^{200}=y\Rightarrow y^{199}=1\Rightarrow y=1\)
c) \(\left(3x-1\right)^{10}=\left(3y-1\right)^{20}\Rightarrow1=\left(3y-1\right)^{10}\)
\(\Rightarrow3y-1=1\) hoặc \(3y-1=-1\)
\(\Rightarrow3y=2\) hoặc \(3y=0\)
\(\Rightarrow y=\frac{2}{3}\) hoặc y =0
d) \(y^{2008}=y^{2010}\Rightarrow1=y^2\Rightarrow\orbr{\begin{cases}y=-1\\y=1\end{cases}}\)
e) \(\left(x-5\right)^2=\left(1-3x\right)^2\Rightarrow x-5=1-3x\Rightarrow x+3x=1+5\Rightarrow4x=6\Rightarrow x=\frac{6}{4}=\frac{3}{2}\)
Can you k for me ,Nhàn Nguyễn Thị Thanh!