x2-\(\sqrt{x+5}\)=5
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27 tháng 9 2021
\(\left(x^2-x-6\right)\left(x^2-5\right)=0\)
\(\Rightarrow\left(x-3\right)\left(x+2\right)\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\\x=\sqrt{5}\\x=-\sqrt{5}\end{matrix}\right.\)
Mà \(x\in Q\)
\(\Rightarrow x=\left\{-2;3\right\}\)
27 tháng 9 2021
Pt\(\Leftrightarrow\)\(\left[{}\begin{matrix}x^2-x-6=0\\x^2-5=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}3\\-2\\-\sqrt{5}\\\sqrt{5}\end{matrix}\right.\)
Đáp án A
Y
24 tháng 6 2023
\(\left(2-\sqrt{5}\right)x^2+\left(6-\sqrt{5}\right)x-8+2\sqrt{5}=0\)
\(\Leftrightarrow\left(2-\sqrt{5}\right)x^2-\left(2-\sqrt{5}\right)x+\left(8-2\sqrt{5}\right)x-(8-2\sqrt{5})=0\)
\(\Leftrightarrow\left(2-\sqrt{5}\right)x\left(x-1\right)+\left(8-2\sqrt{5}\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[\left(2-\sqrt{5}\right)x+\left(8-2\sqrt{5}\right)\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\left(2-\sqrt{5}\right)x=-8+2\sqrt{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-8+2\sqrt{5}}{2-\sqrt{5}}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=6+4\sqrt{5}\end{matrix}\right.\)
Vậy \(S=\left\{1;6+4\sqrt{5}\right\}\)
N
2
NH
11 tháng 7 2023
a)
\(x^2-4\sqrt{15}x+19=0\\ < =>x^2-4\sqrt{15}x+60-41=0\\ < =>\left(x-2\sqrt{15}\right)^2-41=0\\ < =>\left(x-2\sqrt{15}-\sqrt{41}\right)\left(x-2\sqrt{15}+\sqrt{41}\right)=0\\ < =>\left[{}\begin{matrix}x-2\sqrt{15}-\sqrt{41}=0\\x-2\sqrt{15}+\sqrt{41}=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=2\sqrt{15}+\sqrt{41}\\x=2\sqrt{15}-\sqrt{41}\end{matrix}\right.\)
b)
\(4x^2+4\sqrt{5}x+5=0\\ < =>\left(2x+\sqrt{5}\right)^2=0\\ < =>2x+\sqrt{5}=0\\ < =>2x=-\sqrt{5}\\ < =>-\dfrac{\sqrt{5}}{2}\)
11 tháng 7 2023
a: Δ=(4căn 15)^2-4*1*19=164>0
Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x=\dfrac{4\sqrt{5}-2\sqrt{41}}{2}=2\sqrt{5}-\sqrt{41}\\x_2=2\sqrt{5}+\sqrt{41}\end{matrix}\right.\)
b: \(\Leftrightarrow\left(2x\right)^2+2\cdot2x\cdot\sqrt{5}+5=0\)
=>(2x+căn 5)^2=0
=>2x+căn 5=0
=>x=-1/2*căn 5
14 tháng 10 2021
\(a,ĐK:\left\{{}\begin{matrix}x\ge5\\x\le3\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
Vậy pt vô nghiệm
\(b,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=2-5x\\ \Leftrightarrow0x=2\Leftrightarrow x\in\varnothing\)
\(c,ĐK:x\ge-\dfrac{3}{2}\\ PT\Leftrightarrow x^2+4x+5-2\sqrt{2x+3}=0\\ \Leftrightarrow\left(2x+3-2\sqrt{2x+3}+1\right)+\left(x^2+2x+1\right)=0\\ \Leftrightarrow\left(\sqrt{2x+3}-1\right)^2+\left(x+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x+3=1\\x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\left(tm\right)\\ d,PT\Leftrightarrow\left|x-1\right|=\left|2x-1\right|\Leftrightarrow\left[{}\begin{matrix}x-1=2x-1\\x-1=1-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)
ND
Nguyễn Đức Trí
VIP
8 tháng 9 2023
Các đơn thức là :
\(\left(1-\dfrac{1}{\sqrt[]{3}}\right)x^2;x^2.\dfrac{7}{2}\)
YT
1
16 tháng 10 2023
`#3107.101107`
`1/2x + 4/5 = 2x - 8/5`
`=> 1/2x - 2x = -4/5 - 8/5`
`=> -3/2x = -12/5`
`=> x = -12/5 \div (-3/2)`
`=> x = 8/5`
Vậy, `x = 8/5`
_____
`\sqrt{x} = 5`
`=> x = 5^2`
`=> x = 25`
Vậy, `x = 25`
___
`x^2 = 3`
`=> x^2 = (+-\sqrt{3})^2`
`=> x = +- \sqrt{3}`
Vậy, `x \in {-\sqrt{3}; \sqrt{3}}.`
26 tháng 11 2021
\(x^2=1\Rightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
\(x^2=3\Rightarrow\left[{}\begin{matrix}x=-\sqrt{3}\\x=\sqrt{3}\end{matrix}\right.\)
\(x^2=5\Rightarrow\left[{}\begin{matrix}x=-\sqrt{5}\\x=\sqrt{5}\end{matrix}\right.\Rightarrow x=-\sqrt{5}\left(vì.x< 0\right)\)
\(x^2=7\Rightarrow\left[{}\begin{matrix}x=-\sqrt{7}\\x=\sqrt{7}\end{matrix}\right.\Rightarrow x=-\sqrt{7}\left(vì.x< 0\right)\)
\(x^2=9\Rightarrow\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\)
\(\left(x-2\right)^2=2\Rightarrow\left[{}\begin{matrix}x-2=-\sqrt{2}\\x-2=\sqrt{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2-\sqrt{2}\\x=2+\sqrt{2}\end{matrix}\right.\)
\(\left(x-4\right)^2=4\Rightarrow\left[{}\begin{matrix}x-2=-2\\x-2=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
\(\left(x-6\right)^2=6\Rightarrow\left[{}\begin{matrix}x-6=-\sqrt{6}\\x-6=\sqrt{6}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6-\sqrt{6}\\x=6+\sqrt{6}\end{matrix}\right.\)
\(\left(x-8\right)^2=8\Rightarrow\left[{}\begin{matrix}x-8=-2\sqrt{2}\\x-8=2\sqrt{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8-2\sqrt{2}\\x=2+2\sqrt{2}\end{matrix}\right.\)
\(\left(x-10\right)^2=10\Rightarrow\left[{}\begin{matrix}x-10=-\sqrt{10}\\x-10=\sqrt{10}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=10-\sqrt{10}\\x=10+\sqrt{10}\end{matrix}\right.\)
\(\left(x-\sqrt{3}\right)^2=3\Rightarrow\left[{}\begin{matrix}x-\sqrt{3}=-\sqrt{3}\\x-\sqrt{3}=\sqrt{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=2\sqrt{3}\end{matrix}\right.\)
\(\left(x-\sqrt{5}\right)^2=5\Rightarrow\left[{}\begin{matrix}x-\sqrt{5}=-\sqrt{5}\\x-\sqrt{5}=\sqrt{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=2\sqrt{5}\end{matrix}\right.\)