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\(\left|x-1\right|+\left|y+2\right|+\left|z-3\right|=0\)
Ta có: \(\hept{\begin{cases}\left|x-1\right|\ge0\forall x\\\left|y+2\right|\ge0\forall x\\\left|z-3\right|\ge0\forall x\end{cases}\Rightarrow\left|x-1\right|+\left|y+2\right|+\left|z-3\right|\ge0\forall x;y;z}\)
Mà \(\left|x-1\right|+\left|y+2\right|+\left|z-3\right|=0\)
\(\hept{\begin{cases}\left|x-1\right|=0\\\left|y+2\right|=0\\\left|z-3\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-2\\z=3\end{cases}}\)
Vậy \(x=1;y=-2;z=3\)
Fudo lm thiếu 1 trường hợp r
Ta có \(\left(2x-5\right)^2=\left|2x-5\right|\)
\(\Leftrightarrow\orbr{\begin{cases}\left(2x-5\right)^2=2x-5\\\left(2x-5^2\right)=5-2x\end{cases}}\)
TH1: \(\left(2x-5\right)^2=2x-5\)
\(\Leftrightarrow\left(2x-5\right)^2-\left(2x-5\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x-5-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-5=0\\2x-5-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=5\\2x-6=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\2x=6\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=3\end{cases}}\) (1)
TH2: \(\left(2x-5\right)^2=5-2x\)
\(\Leftrightarrow\left(2x-5\right)^2-\left(5-2x\right)=0\)
\(\Leftrightarrow\left(2x-5\right)^2+2x-5=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x-5+1\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-5=0\\2x-4=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=5\\2x=4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=2\end{cases}}\) (2)
Từ (1) và (2) \(\Leftrightarrow x\in\left\{\frac{5}{2};2;3\right\}\)
Vậy \(x\in\left\{\frac{5}{2};2;3\right\}\)
@@ Học tốt
\(\left(x-2\right)^8=\left(x-2\right)^6\)
\(\Leftrightarrow\left(x-2\right)^8-\left(x-2\right)^6=0\)
\(\Leftrightarrow\left(x-2\right)^6\left[\left(x-2\right)^2-1\right]=0\)
\(\Leftrightarrow\left(x-2\right)^6\left(x-3\right)\left(x-1\right)=0\)
\(\Rightarrow x=2;x=3;x=1\)
Ta có : |2x - 1| + 1 = x
=> |2x - 1| = x - 1
\(\Leftrightarrow\orbr{\begin{cases}2x-1=x-1\\2x-1=1-x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-x=-1+1\\2x+x=1+1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\3x=2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{2}{3}\end{cases}}\)
\(a,\frac{1}{2}x+\frac{5}{2}=\frac{7}{2}x-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}x+\frac{5}{2}-\frac{7}{2}x=-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}x-\frac{7}{2}x+\frac{5}{2}=-\frac{3}{4}\)
\(\Leftrightarrow-3x+\frac{5}{2}=-\frac{3}{4}\)
\(\Leftrightarrow-3x=-\frac{13}{4}\)
\(\Leftrightarrow x=-\frac{13}{4}:(-3)=-\frac{13}{4}:\frac{-3}{1}=-\frac{13}{4}\cdot\frac{-1}{3}=\frac{13}{12}\)
\(b,\frac{2}{3}x-\frac{2}{5}=\frac{1}{2}x-\frac{1}{3}\)
\(\Leftrightarrow\frac{2}{3}x-\frac{2}{5}-\frac{1}{2}x=-\frac{1}{3}\)
\(\Leftrightarrow\frac{2}{3}x-\frac{1}{2}x-\frac{2}{5}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{6}x-\frac{2}{5}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{6}x=\frac{1}{15}\)
\(\Leftrightarrow x=\frac{1}{15}:\frac{1}{6}=\frac{1}{15}\cdot6=\frac{6}{15}=\frac{2}{5}\)
\(c,\frac{1}{3}x+\frac{2}{5}(x+1)=0\)
\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)
\(\Leftrightarrow\frac{11}{15}x=-\frac{2}{5}\)
\(\Leftrightarrow x=-\frac{6}{11}\)
d,e,f Tương tự
Thay x =2 ; y= -1 vào biểu thức ta có:
16.2.(-1)^5 - 2.2^3.(-1)
=16.2.(-1) - 2.1/8
=-32 - 1/4
=-129/4
vậy...........................
học tốt!
