1.Tìm x,biết: [x]+[x-1]+[x-2]=4x-8
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NT
1
S
19 tháng 2 2018
Ta thấy \(\left|x+\frac{1}{8}\right|\ge0\forall x;\left|x+\frac{2}{8}\right|\ge0\forall x;\left|x+\frac{5}{8}\right|\ge0\forall x\)
\(\Rightarrow\left|x+\frac{1}{8}\right|+\left|x+\frac{2}{8}\right|+\left|x+\frac{5}{8}\right|\ge0\)
\(\Rightarrow4x\ge0\Rightarrow x\ge0\)
\(\Rightarrow x+\frac{1}{8}+x+\frac{2}{8}+x+\frac{5}{8}=4x\)
\(\Rightarrow3x+1=4x\)
=> x = 1 (t/m)
Vậy x=1
29 tháng 11 2023
a: \(x^3-4x^2-x+4=0\)
=>\(\left(x^3-4x^2\right)-\left(x-4\right)=0\)
=>\(x^2\left(x-4\right)-\left(x-4\right)=0\)
=>\(\left(x-4\right)\left(x^2-1\right)=0\)
=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)
b: Sửa đề: \(x^3+3x^2+3x+1=0\)
=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)
=>\(\left(x+1\right)^3=0\)
=>x+1=0
=>x=-1
c: \(x^3+3x^2-4x-12=0\)
=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)
=>\(x^2\cdot\left(x+3\right)-4\left(x+3\right)=0\)
=>\(\left(x+3\right)\left(x^2-4\right)=0\)
=>\(\left(x+3\right)\left(x-2\right)\left(x+2\right)=0\)
=>\(\left[{}\begin{matrix}x+3=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\\x=-2\end{matrix}\right.\)
d: \(\left(x-2\right)^2-4x+8=0\)
=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)
=>\(\left(x-2\right)^2-4\left(x-2\right)=0\)
=>\(\left(x-2\right)\left(x-2-4\right)=0\)
=>(x-2)(x-6)=0
=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)
23 tháng 8 2021
3) \(x\left(x-4\right)+\left(x-4\right)^2=0\Leftrightarrow\left(x-4\right)\left(x+x-4\right)=0\Leftrightarrow2\left(x-4\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
BT
0
NO
2
8 tháng 12 2017
a) \(\left(x-1\right)\left(2x+3\right)-x\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+3-x\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}}\)
Vậy \(x=1;-3\)
b) \(x^2-4x+8=2x-1\)
\(\Leftrightarrow x^2-4x+8-2x+1=0\)
\(\Leftrightarrow x^2-6x+9=0\)
\(\Leftrightarrow\left(x-3\right)^2=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)
Vậy x=3
9 tháng 12 2017
a) \(\left(x-1\right)\left(2x+3\right)-x\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+3-x\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}}\)
Vậy \(x=1;-3\)
b) \(x^2-4x+8=2x-1\)
\(\Leftrightarrow x^2-4x+8-2x+1=0\)
\(\Leftrightarrow x^2-6x+9=0\)
\(\Leftrightarrow\left(x-3\right)^2=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)
Vậy \(x=3\)
AH
2
5 tháng 8 2015
\(\Rightarrow\frac{3}{4}x-\frac{1}{4}=2x-6+\frac{1}{4}x\)
\(\Rightarrow\frac{3}{4}x-2x-\frac{1}{4}x=-6+\frac{1}{4}\)
\(\Rightarrow-\frac{3}{2}x=-\frac{23}{4}\)
\(\Rightarrow x=\frac{23}{4}:\frac{3}{2}=\frac{23}{6}\)
\(\Rightarrow\frac{8}{9}x-\frac{1}{3}x=\frac{2}{3}+\frac{11}{3}\)
\(\Rightarrow\frac{5}{9}x=\frac{13}{3}\)
\(\Rightarrow x=\frac{13}{3}:\frac{5}{9}=\frac{39}{5}\)
30 tháng 8 2016
\(\Rightarrow\frac{3}{4}x-\frac{1}{4}=2x-6+\frac{1}{4}x\)
\(\Rightarrow\frac{3}{4}x-2x-\frac{1}{4}x=-6+\frac{1}{4}\)
\(\Rightarrow-\frac{3}{2}x=-\frac{23}{4}\)
\(\Rightarrow x=\frac{23}{4}:\frac{3}{2}=\frac{23}{6}\)
\(\Rightarrow\frac{8}{9}x-\frac{1}{3}x=\frac{2}{3}+\frac{11}{3}\)
\(\Rightarrow\frac{5}{9}x=\frac{13}{3}\)
\(\Rightarrow x=\frac{13}{3}:\frac{5}{9}=\frac{39}{5}\)