\(\sqrt{x-7}\) + \(\sqrt{9-x}\) = \(x^2\)-16x+66
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14 tháng 11 2019
\(\frac{x}{16}=\frac{2\sqrt{27+7\sqrt{5}}}{\sqrt{10}+7\sqrt{2}}\)
\(\Leftrightarrow\frac{x}{16}=\frac{\sqrt{54+14\sqrt{5}}}{\sqrt{5}+7}\)
\(\Leftrightarrow\frac{x}{16}=\frac{\sqrt{49+2.7.\sqrt{5}+5}}{\sqrt{5}+7}\)
\(\Leftrightarrow\frac{x}{16}=\frac{\sqrt{\left(\sqrt{5}+7\right)^2}}{\sqrt{5}+7}=\frac{\sqrt{5}+7}{\sqrt{5}+7}=1\)
\(\Leftrightarrow x=16\)
30 tháng 9 2021
\(\sqrt{\dfrac{x^2+2x+1}{16x^2}}=\sqrt{\dfrac{\left(x+1\right)^2}{16x^2}}=\dfrac{\left|x+1\right|}{4\left|x\right|}=\dfrac{1-x}{-4x}=\dfrac{x-1}{4x}\left(do.x\le-1\right)\)
30 tháng 8 2018
\(VT\)
\(A=\sqrt{x-7}+\sqrt{9-x}\)
\(\Rightarrow A^2=2+2\sqrt{\left(x-7\right)\left(9-x\right)}\le2+\left(x-7\right)+\left(9-x\right)=4\)
\(\Rightarrow A\le2\)
\(VP\)
\(B=\left(x-8\right)^2+2\ge2\)
Theo đề bài , \(A=B\Rightarrow A=B=2\)
Do đó \(x-7=9-x\Leftrightarrow x=8\)
Vậy \(x=8\)
P/s tham khảo nha
24 tháng 9 2021
\(P=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+1:\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{2\sqrt{x}+7}{x-4}\right)\)
\(=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+1:\left(\dfrac{x+2\sqrt{x}-x+\sqrt{x}+2-2\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\)
\(=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-5}\)
\(=\dfrac{-x+8\sqrt{x}-15+\left(x-4\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{-x+8\sqrt{x}-15+x\sqrt{x}-2x-4\sqrt{x}+8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{x\sqrt{x}-3x+4\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\)
24 tháng 9 2021
\(ĐK:x\ge0;x\ne4\\ P=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+1:\dfrac{x+2\sqrt{x}-x+\sqrt{x}+2-2\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ P=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-5}\\ P=\dfrac{\left(3-\sqrt{x}\right)\left(\sqrt{x}-5\right)+\left(x-4\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\\ P=\dfrac{8\sqrt{x}-15-x+x\sqrt{x}-2x-4\sqrt{x}+8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\\ P=\dfrac{x\sqrt{x}-3x+4\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\)
BC
1
BC
0
4 tháng 9 2021
g: \(\dfrac{\sqrt{x}+3}{x\sqrt{x}+27}=\dfrac{1}{x-3\sqrt{x}+9}\)
h: \(\dfrac{2x-2\sqrt{x}+2}{x\sqrt{x}+1}=\dfrac{2}{\sqrt{x}+1}\)
i: \(\dfrac{x-3\sqrt{x}+2}{x-\sqrt{x}}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
k: \(\dfrac{x+7\sqrt{x}+12}{x-9}=\dfrac{\sqrt{x}+4}{\sqrt{x}-3}\)
i: \(\dfrac{x+\sqrt{x}-2}{x-2\sqrt{x}+1}=\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\)
HD
26 tháng 7 2021
Bài 2
b, `\sqrt{3x^2}=x+2` ĐKXĐ : `x>=0`
`=>(\sqrt{3x^2})^2=(x+2)^2`
`=>3x^2=x^2+4x+4`
`=>3x^2-x^2-4x-4=0`
`=>2x^2-4x-4=0`
`=>x^2-2x-2=0`
`=>(x^2-2x+1)-3=0`
`=>(x-1)^2=3`
`=>(x-1)^2=(\pm \sqrt{3})^2`
`=>` $\left[\begin{matrix} x-1=\sqrt{3}\\ x-1=-\sqrt{3}\end{matrix}\right.$
`=>` $\left[\begin{matrix} x=1+\sqrt{3}\\ x=1-\sqrt{3}\end{matrix}\right.$
Vậy `S={1+\sqrt{3};1-\sqrt{3}}`
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 9 2021
ĐKXĐ: \(x\ge0;x\ne3\)
\(B=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{-3\sqrt{x}-3}{x-9}\)
Ta có: \(\sqrt{x-7}\le\frac{x-7+1}{2}=\frac{x-6}{2}\)(bđt cosi)
\(\sqrt{9-x}\le\frac{9-x+1}{2}=\frac{10-x}{2}\)
=> \(VT=\sqrt{x-7}+\sqrt{9-x}\le\frac{x-6}{2}+\frac{10-x}{2}=\frac{x-6+10-x}{2}=2\)
\(VP=x^2-16x+66=\left(x-8\right)^2+2\ge2\)
=> \(VT=VP\Leftrightarrow\hept{\begin{cases}x-7=1\\9-x=1\\x-8=0\end{cases}}\) <=> x = 8
Vậy S = {8}
\(\sqrt{x-7}+\sqrt{9-x}=x^2-16x+66\left(7\le x\le9\right)\)
Đặt \(A=\sqrt{x-7}+\sqrt{9-x}\)
\(\Rightarrow A^2=\left(\sqrt{x-7}+\sqrt{9-x}\right)^2\)
Áp dụng bất đảng thức Bunhiacopxki ta có:
\(\left(\sqrt{x-7}+\sqrt{9-x}\right)^2\le\left(x-7+9-x\right)\left(1+1\right)=4\)
=> \(A\le2\)
Ta có: \(x^2-16x+66=\left(x-8\right)^2+2\ge2\)
Dấu = xảy ra
\(\Leftrightarrow\hept{\begin{cases}\frac{\sqrt{x-7}}{1}=\frac{\sqrt{9-x}}{1}\\x-8=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{x-7}=\sqrt{9-x}\\x=8\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x-7=9-x\\x=8\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=8\\x=8\end{cases}\left(tm\right)}\)
Vậy x = 8