2√x-9/x-5√x+6 - √x +3/√x -2√-2√x+1/3-√x
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19 tháng 6 2023
`@` `\text {Ans}`
`\downarrow`
`2+(x+3)=7`
`\Rightarrow x+3=7-2`
`\Rightarrow x+3=5`
`\Rightarrow x=5-3`
`\Rightarrow x=2`
`5+(3+x)=10`
`\Rightarrow 3+x=10-5`
`\Rightarrow 3+x=5`
`\Rightarrow x=5-3`
`\Rightarrow x=2`
`(4+x)+1=7`
`\Rightarrow 4+x=7-1`
`\Rightarrow 4+x=6`
`\Rightarrow x=6-4`
`\Rightarrow x=2`
`(x+5)+3=9`
`\Rightarrow x+5=9-3`
`\Rightarrow x+5=6`
`\Rightarrow x=6-5`
`\Rightarrow x=1`
`(x-1)-4=7`
`\Rightarrow x-1=7+4`
`\Rightarrow x-1=11`
`\Rightarrow x=11+1`
`\Rightarrow x=12`
`4-(6-x)=1`
`\Rightarrow 6-x=4-1`
`\Rightarrow 6-x=3`
`\Rightarrow x=6-3`
`\Rightarrow x=3`
H9
HT.Phong (9A5)
CTVHS
19 tháng 6 2023
\(2+\left(x+3\right)=7\)
\(\Rightarrow2+x+3=7\)
\(\Rightarrow x+5=7\)
\(\Rightarrow x=2\)
\(5+\left(3+x\right)=10\)
\(\Rightarrow5+3+x=10\)
\(\Rightarrow x+8=10\)
\(\Rightarrow x=2\)
\(\left(4+x\right)+1=7\)
\(\Rightarrow4+x+1=7\)
\(\Rightarrow x+5=7\)
\(\Rightarrow x=2\)
\(\left(x+5\right)+3=9\)
\(=x+5+3=9\)
\(\Rightarrow x+8=9\)
\(\Rightarrow x=1\)
\(\left(x-1\right)-4=7\)
\(\Rightarrow x-1-4=7\)
\(\Rightarrow x-5=7\)
\(\Rightarrow x=12\)
\(4-\left(6-x\right)=1\)
\(\Rightarrow4-6-x=1\)
\(\Rightarrow-2-x=1\)
\(\Rightarrow x=-3\)
HL
7 tháng 2 2022
a) \(\text{}/3x-5/-\frac{1}{7}=\frac{1}{3}\) b)\(\left(\frac{3}{5}x-\frac{2}{3}x-x\right).\frac{1}{7}=\frac{-5}{21}\)
\(/3x-5/=\frac{10}{21}\) \([x.\left(\frac{3}{5}-\frac{2}{3}-1\right)]=\frac{-5}{21}.7\)
\(\Rightarrow3x-5=\frac{10}{21}hay3x-5=\frac{-10}{21}\) \(\left[x.\frac{-16}{15}\right]=\frac{-5}{3}\)
\(3x=\frac{115}{21}\) \(3x=\frac{95}{21}\) \(x=\frac{25}{16}\)
\(x=\frac{115}{63}\) \(x=\frac{95}{63}\) Vậy x = \(\frac{25}{16}\)
Vậy x \(\in\left\{\frac{115}{63};\frac{95}{63}\right\}\)
20 tháng 8 2021
\(1,\)
\(2x\left(x-3\right)-\left(3-x\right)=0\)
\(\Leftrightarrow2x\left(x-3\right)+\left(x-3\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=3\end{cases}}\)
\(2,\)
\(3x\left(x+5\right)-6\left(x+5\right)=0\)
\(\Leftrightarrow\left(3x-6\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-6=0\\x+5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)
\(3,\)
\(x^4-x^2=0\)
\(\Leftrightarrow x^2\left(x^2-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x^2-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
\(4,\)
\(x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(5,\)
\(x\left(x+6\right)-10\left(x-6\right)=0\)
\(\Leftrightarrow x^2+6x-10x+60=0\)
\(\Leftrightarrow x^2-4x+60=0\)
\(\Leftrightarrow x^2-4x+4+56=0\)
\(\Leftrightarrow\left(x-2\right)^2=-56\)(Vô lý)
=> Phương trình vô nghiệm
5 tháng 8 2020
Bài 4:
a) Ta có: \(x^9-x^7-x^6-x^5+x^4+x^3+x^2-1\)
\(=\left(x^9-x^7\right)-\left(x^6-x^4\right)-\left(x^5-x^3\right)+\left(x^2-1\right)\)
\(=x^7\left(x^2-1\right)-x^4\left(x^2-1\right)-x^3\left(x^2-1\right)+\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^7-x^4-x^3+1\right)\)
\(=\left(x^2-1\right)\cdot\left[x^4\left(x^3-1\right)-\left(x^3-1\right)\right]\)
\(=\left(x^2-1\right)\cdot\left(x^3-1\right)\cdot\left(x^4-1\right)\)
\(=\left(x-1\right)\left(x+1\right)\cdot\left(x-1\right)\left(x^2+x+1\right)\cdot\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\)
\(=\left(x-1\right)^3\cdot\left(x+1\right)^2\cdot\left(x^2+1\right)\cdot\left(x^2+x+1\right)\)
5 tháng 8 2020
a, Ta có : \(x^5-x^4-x^3-x^2-x-2\)
\(=x^5-2x^4+x^4-2x^3+x^3-2x^2+x^2-2x+x-2\)
\(=x^4\left(x-2\right)+x^3\left(x-2\right)+x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)\)
\(=\left(x-2\right)\left(x^4+x^3+x^2+x+1\right)\)
29 tháng 7 2021
a) Ta có: \(\left(x-3\right)=\left(3-x\right)^2\)
\(\Leftrightarrow\left(x-3\right)^2-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
b) Ta có: \(x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}=\dfrac{1}{64}\)
\(\Leftrightarrow x^3+3\cdot x^2\cdot\dfrac{1}{2}+3\cdot x\cdot\dfrac{1}{4}+\left(\dfrac{1}{2}\right)^3=\dfrac{1}{64}\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^3=\left(\dfrac{1}{4}\right)^3\)
\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\)
hay \(x=-\dfrac{1}{4}\)
c) Ta có: \(8x^3-50x=0\)
\(\Leftrightarrow2x\left(4x^2-25\right)=0\)
\(\Leftrightarrow x\left(2x-5\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)
29 tháng 7 2021
e) Ta có: \(x\left(x+3\right)-x^2-3x=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)
f) Ta có: \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-3\end{matrix}\right.\)