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bài 1 : a,ta có 3/x-1 =4/y-2=5/z-3 => x-1/3=y-2/4=z-3/5
áp dụng .... => x-1+y-2+z-3 / 3+4+5 = x+y+z-1-2-3/3+4+5 = 12/12=1
do x-1/3 = 1 => x-1 = 3 => x= 4 ( tìm y,z tương tự
Bài 1:
a) Ta có: 3/x - 1 = 4/y - 2 = 5/z - 3 => x - 1/3 = y - 2/4 = z - 3/5 áp dụng ... =>x - 1 + y - 2 + z - 3/3 + 4 + 5 = x + y + z - 1 - 2 - 3/3 + 4 + 5 = 12/12 = 1 do x - 1/3 = 1 => x - 1 = 3 => x = 4 ( tìm y, z tương tự )
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(x-1)x+2 = (x-1)x+6
=> (x-1)x+6-(x-1)x+2=0
=> (x-1)x+2[(x-1)4-1]=0
\(\Rightarrow\orbr{\begin{cases}\left(x-1\right)^2=0\\\left(x-1\right)^4-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x-1=0\\\left(x-1\right)^4=1\end{cases}}\)
\(\Rightarrow\)x=1 hoặc \(\Rightarrow\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}\)
Vậy \(x\in\left\{1;2;0\right\}\)
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ta có: \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+6}\Leftrightarrow\left(x-1\right)^{x+6}-\left(x-1\right)^{x+2}=0\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^{x+4}-1\right]=0\)
\(\Leftrightarrow[\begin{cases}\left(x-1\right)^{x+2}=0\\\left(x-1\right)^{x+4}-1=0\end{cases}\Leftrightarrow[\begin{cases}x-1=0\\x-1=1\end{cases}\Leftrightarrow[\begin{cases}x=1\\x=0\end{cases}\)
Vậy x = 1 hoặc x = 0
Học tốt nhé ^3^
(x-1)^(x+2)=(x-1)^(x+6)
[(x-1)^x].(x-1)^2=[(x-1)^x].(x-1)^6
(x-1)^2=(x-1)^6
(X-1)^2=(x-1)^2.(x-1)^4
1=(x-1)^4
1=x-1
1+1=x
2=x
Nhớ k cho mình nha
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\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\)
\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{11-10}{10.11}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\)
\(=1-\frac{1}{11}=\frac{10}{11}\)
Phương trình ban đầu tương đương với:
\(10x+\frac{10}{11}=11x\)
\(\Leftrightarrow x=\frac{10}{11}\)
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Bài 2:
a) Ta có: \(\left|x-2\right|=\left|4-x\right|\)
\(\Leftrightarrow x-2=4-x\)
\(\Leftrightarrow2x=6\)
hay x=3
b) Ta có: \(\left(\left|2x-1\right|-3\right)\cdot\left(-2\right)+\left(-5\right)=6\)
\(\Leftrightarrow\left(\left|2x-1\right|-3\right)\cdot\left(-2\right)=11\)
\(\Leftrightarrow\left|2x-1\right|-3=\dfrac{-11}{2}\)
\(\Leftrightarrow\left|2x-1\right|=\dfrac{-11}{2}+\dfrac{6}{2}=\dfrac{-5}{2}\)(Vô lý)
\(\left(x-1\right)^2=\left(x-1\right)^6\Leftrightarrow\left(x-1\right)^2-\left(x-1\right)^6=0\)
\(\Leftrightarrow\left(x-1\right)^2\left[1-\left(x-1\right)^4\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)^2=0\\\left(x-1\right)^4=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x-1=1;x-1=-1\end{cases}}\)
\(\Leftrightarrow x=0;1;2\)