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\(ĐK:x\ge1\\ PT\Leftrightarrow12\sqrt{x-1}-\sqrt{x-1}-8\sqrt{x-1}+\sqrt{x-1}=16\\ \Leftrightarrow4\sqrt{x-1}=16\\ \Leftrightarrow\sqrt{x-1}=4\\ \Leftrightarrow x-1=16\\ \Leftrightarrow x=17\left(tm\right)\)
`2sqrt{36x-36}-1/3sqrt{9x-9}-4sqrt{4x-4}+sqrt{x-1}=16`
`ĐK:x>=1`
`pt<=>2sqrt{36(x-1)}-1/3sqrt{9(x-1)}-4sqrt{4(x-1)}+sqrt{x-1}=16`
`<=>12sqrt{x-1}-sqrt{x-1}-8sqrt{x-1}+sqrt{x-1}=16`
`<=>4sqrt{x-1}=16`
`<=>sqrt{x-1}=4`
`<=>x-1=16`
`<=>x=17(tmđk)`
Vậy `S={17}`
\(a,\sqrt{x+1}=\sqrt{2-x}\)
\(\Rightarrow x+1=2-x\)
\(\Rightarrow2x=1\)
\(\Rightarrow x=\frac{1}{2}\)
a) \(ĐKXĐ:-1\le x\le2\)
Bình phương 2 vế ta có:
\(x+1=2-x\)\(\Leftrightarrow2x=1\)\(\Leftrightarrow x=\frac{1}{2}\)( đpcm )
Vậy \(x=\frac{1}{2}\)
b) \(ĐKXĐ:x\ge1\)
\(\sqrt{36x-36}-\sqrt{9x-9}-\sqrt{4x-4}=16-\sqrt{x-1}\)
\(\Leftrightarrow\sqrt{36\left(x-1\right)}-\sqrt{9\left(x-1\right)}-\sqrt{4\left(x-1\right)}+\sqrt{x-1}=16\)
\(\Leftrightarrow6\sqrt{x-1}-3\sqrt{x-1}-2\sqrt{x-1}+\sqrt{x-1}=16\)
\(\Leftrightarrow2\sqrt{x-1}=16\)\(\Leftrightarrow\sqrt{x-1}=8\)
\(\Leftrightarrow x-1=64\)\(\Leftrightarrow x=65\)( thỏa mãn ĐKXĐ )
Vậy \(x=65\)
c) \(ĐKXĐ:x\ge1\)
\(\sqrt{16x-16}-\sqrt{9x-9}+\sqrt{4x-4}+\sqrt{x-1}=8\)
\(\Leftrightarrow\sqrt{16\left(x-1\right)}-\sqrt{9\left(x-1\right)}+\sqrt{4\left(x-1\right)}+\sqrt{x-1}=8\)
\(\Leftrightarrow4\sqrt{x-1}-3\sqrt{x-1}+2\sqrt{x-1}+\sqrt{x-1}=8\)
\(\Leftrightarrow4\sqrt{x-1}=8\)\(\Leftrightarrow\sqrt{x-1}=2\)
\(\Leftrightarrow x-1=4\)\(\Leftrightarrow x=5\)( thỏa mãn ĐKXĐ )
Vậy \(x=5\)
( với x
0; x
1)
. Tính độ dài HD và diện tích tam giác AHD
Bài 5:
\(\widehat{B}=60^0\)
\(AB=8\sqrt{3}\left(cm\right)\)
\(BC=16\sqrt{3}\left(cm\right)\)
a,\(\sqrt{\left(3x-1\right)^2}=5=>|3x-1|=5=>\left[{}\begin{matrix}3x-1=5\\3x-1=-5\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)
b, \(\sqrt{4x^2-4x+1}=3=\sqrt{\left(2x-1\right)^2}=3=>\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
c, \(\sqrt{x^2-6x+9}+3x=4=>|x-3|=4-3x\)
TH1: \(|x-3|=x-3< =>x\ge3=>x-3=4-3x=>x=1,75\left(ktm\right)\)
TH2 \(|x-3|=3-x< =>x< 3=>3-x=4-3x=>x=0,5\left(tm\right)\)
Vậy x=0,5...
d, đk \(x\ge-1\)
=>pt đã cho \(< =>9\sqrt{x+1}-6\sqrt{x+1}+4\sqrt{x+1}=12\)
\(=>7\sqrt{x+1}=12=>x+1=\dfrac{144}{49}=>x=\dfrac{95}{49}\left(tm\right)\)
a) Ta có: \(\sqrt{\left(3x-1\right)^2}=5\)
\(\Leftrightarrow\left|3x-1\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=5\\3x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=6\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)
b) Ta có: \(\sqrt{4x^2-4x+1}=3\)
\(\Leftrightarrow\left|2x-1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=4\\2x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
c) Ta có: \(\sqrt{x^2-6x+9}+3x=4\)
\(\Leftrightarrow\left|x-3\right|=4-3x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=4-23x\left(x\ge3\right)\\x-3=23x-4\left(x< 3\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+23x=4+3\\x-23x=4+3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{24}\left(loại\right)\\x=\dfrac{-4}{22}=\dfrac{-2}{11}\left(loại\right)\end{matrix}\right.\)
2, \(\dfrac{1}{2}\sqrt{x+1}-\dfrac{3}{2}\sqrt{9x+9}+24\sqrt{\dfrac{x+1}{64}}=-17\)
\(\Leftrightarrow\dfrac{1}{2}\sqrt{x+1}-\dfrac{9}{2}\sqrt{x+1}+3\sqrt{x+1}=-17\)
\(\Leftrightarrow-\sqrt{x+1}=-17\)
\(\Leftrightarrow x+1=289\left(x>0\right)\)
\(\Leftrightarrow x=288\)
Vậy x = 288
3, \(-5x+7\sqrt{x}+12=0\)
\(\Leftrightarrow-5x+12\sqrt{x}-5\sqrt{x}+12=0\)
\(\Leftrightarrow\sqrt{x}\left(12-5\sqrt{x}\right)+\left(12-5\sqrt{x}\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(12-5\sqrt{x}\right)=0\)
Do \(\sqrt{x}+1>0\)
\(\Rightarrow12-5\sqrt{x}=0\Leftrightarrow x=\dfrac{144}{25}\)
Vậy...
1. (Đề có chút sai sai nên mình sửa lại nhé) \(\sqrt{36x-36}-\sqrt{9x-9}-\sqrt{4x-4}=16-\sqrt{x-1}\)
(ĐK: \(x\ge1\))
\(\Leftrightarrow\sqrt{36\left(x-1\right)}-\sqrt{9\left(x-1\right)}-\sqrt{4\left(x-1\right)}=16-\sqrt{x-1}\)
\(\Leftrightarrow6\sqrt{x-1}-3\sqrt{x-1}-2\sqrt{x-1}+\sqrt{x-1}=16\)
\(\Leftrightarrow2\sqrt{x-1}=16\)
\(\Leftrightarrow\sqrt{x-1}=8\)
\(\Leftrightarrow x-1=64\)
\(\Leftrightarrow x=65\left(tm\right)\)
Vậy pt đã cho có nghiệm x=65.
2. \(\dfrac{1}{2}\sqrt{x+1}-\dfrac{3}{2}\sqrt{9x+9}+24\sqrt{\dfrac{x+1}{64}}=-17\)
(ĐK: \(x\ge-1\))
\(\Leftrightarrow\dfrac{1}{2}\sqrt{x+1}-\dfrac{3}{2}\sqrt{9\left(x+1\right)}+3\sqrt{x+1}=-17\)
\(\Leftrightarrow\dfrac{1}{2}\sqrt{x+1}-\dfrac{9}{2}\sqrt{x+1}+3\sqrt{x+1}=-17\)
\(\Leftrightarrow-\sqrt{x+1}=-17\)
\(\Leftrightarrow\sqrt{x+1}=17\)
\(\Leftrightarrow x+1=289\)
\(\Leftrightarrow x=288\left(tm\right)\)
Vậy \(S=\left\{288\right\}\)
3. \(-5x+7\sqrt{x}+12=0\) (ĐK: \(x\ge0\))
\(\Leftrightarrow5x-7\sqrt{x}-12=0\)
\(\Leftrightarrow5x+5\sqrt{x}-12\sqrt{x}-12=0\)
\(\Leftrightarrow5\sqrt{x}\left(\sqrt{x}+1\right)-12\left(\sqrt{x}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(5\sqrt{x}-12\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+1=0\\5\sqrt{x}-12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=-1\left(vô.lý\right)\\5\sqrt{x}=12\end{matrix}\right.\Leftrightarrow\sqrt{x}=\dfrac{12}{5}\Leftrightarrow x=\dfrac{144}{25}\left(tm\right)\)
Vậy pt có nghiệm \(x=\dfrac{144}{25}\)
Sửa đề: \(2\sqrt{36x-36}-\dfrac{1}{3}\sqrt{9x-9}-4\sqrt{4x-4}+\sqrt{x-1}=16\)
\(\Leftrightarrow12\sqrt{x-1}-\sqrt{x-1}-8\sqrt{x-1}+\sqrt{x-1}=16\)
=>4 căn x-1=16
=>căn x-1=4
=>x-1=16
=>x=17