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a) Ta có: \(\sqrt{4x-8}+5\sqrt{x-2}-\sqrt{9x-18}=20\) \(\left(ĐK:x\ge2\right)\)
\(\Leftrightarrow\sqrt{4}.\sqrt{x-2}+5\sqrt{x-2}-\sqrt{9}.\sqrt{x-2}=20\)
\(\Leftrightarrow2.\sqrt{x-2}+5\sqrt{x-2}-3.\sqrt{x-2}=20\)
\(\Leftrightarrow4.\sqrt{x-2}=20\)
\(\Leftrightarrow\sqrt{x-2}=5\)
\(\Leftrightarrow x-2=25\)
\(\Leftrightarrow x=27\left(TM\right)\)
Vậy \(S=\left\{27\right\}\)
2/ \(\Rightarrow5\sqrt{x+1}-6\sqrt{x+1}+3\sqrt{x+1}=2\sqrt{2x+3}\)
\(\Rightarrow\sqrt{x+1}\left(5-6+3\right)=2\sqrt{2x+3}\)
\(\Rightarrow2\sqrt{x+1}=2\sqrt{2x-3}\Rightarrow\sqrt{x+1}=\sqrt{2x+3}\)
\(\Rightarrow x+1=2x+3\Rightarrow x=-2\)
bài 1:
đkxđ: x\(\ge\)0;y\(\ge\)1;z\(\ge\)2
\(\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}=\frac{1}{2}\left(x+y+z\right)\)
\(\Leftrightarrow2\left(\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}\right)=2.\frac{1}{2}\left(x+y+z\right)\)
\(\Leftrightarrow2\sqrt{x}+2\sqrt{y-1}+2\sqrt{z-2}=x+y+z\)
\(\Leftrightarrow x-2\sqrt{x}+y-2\sqrt{y-1}+z-2\sqrt{z-2}=0\)
\(\Leftrightarrow x-2\sqrt{x}+1+y-1-2\sqrt{y-1}+1+z-2-2\sqrt{z-2}+1+1=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y-1}-1\right)^2+\left(\sqrt{z-2}-1\right)^2=-1\)(Vô lí)
Vậy phương trình vô nghiệm
bài 2:
đkxđ: x+1\(\ne\)0
<=>x\(\ne\)-1
\(5\sqrt{x+1}-\sqrt{36x+36}+\sqrt{9x+9}=\sqrt{8x+12}\)
\(\Leftrightarrow5\sqrt{x+1}-\sqrt{36.\left(x+1\right)}+\sqrt{9.\left(x+1\right)}=\sqrt{8x+12}\)
\(\Leftrightarrow5\sqrt{x+1}-6\sqrt{x+1}+3\sqrt{x+1}=\sqrt{8x+12}\)
\(\Leftrightarrow2\sqrt{x+1}=\sqrt{8x+12}\)
\(\Leftrightarrow4.\left(x+1\right)=8x+12\)
\(\Leftrightarrow4x+4=8x+12\)
\(\Leftrightarrow-4x=8\)
\(\Leftrightarrow x=-2\)(thõa mãn)
Vậy x=-2
Bài 1:
a: \(=\left|5-\sqrt{3}\right|-\left|\sqrt{3}-2\right|\)
\(=5-\sqrt{3}-2+\sqrt{3}=3\)
b; \(B=\dfrac{\left(2-\sqrt{3}\right)\cdot\sqrt{52+30\sqrt{3}}-\left(2+\sqrt{3}\right)\cdot\sqrt{52-30\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\left(2-\sqrt{3}\right)\cdot\left(3\sqrt{3}+5\right)-\left(2+\sqrt{3}\right)\left(3\sqrt{3}-5\right)}{\sqrt{2}}\)
\(=\dfrac{6\sqrt{3}+10-9-5\sqrt{3}-6\sqrt{3}+10-9+5\sqrt{3}}{\sqrt{2}}\)
\(=\dfrac{20-18}{\sqrt{2}}=\sqrt{2}\)
c: \(C=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3+3-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)}=1\)
d: \(A=\left(\sqrt{5}-1\right)\cdot\sqrt{6+2\sqrt{5}}\)
\(=\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)=5-1=4\)
\(a,\sqrt{x+1}=\sqrt{2-x}\)
\(\Rightarrow x+1=2-x\)
\(\Rightarrow2x=1\)
\(\Rightarrow x=\frac{1}{2}\)
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 \(x=5\)
\(\sqrt{36x-36}-\sqrt{9x-9}-\sqrt{4x-4}=16-\sqrt{x-1}\)
\(\sqrt{36\left(x-1\right)}-\sqrt{9\left(x-1\right)}-\sqrt{4\left(x-1\right)}=16-\sqrt{x-1}\)
\(6\sqrt{x-1}-3\sqrt{x-1}-2\sqrt{x-1}+\sqrt{x-1}=16\)
\(\sqrt{x-1}\left(6-3-2+1\right)=16\)
\(2\sqrt{x-1}=16\)
\(\sqrt{x-1}=8\)
\(\left(\sqrt{x-1}\right)^2=8^2\)
\(x-1=64\)
\(x=64+1=65\)
\(\sqrt{36x-36}-\sqrt{9x-9}-\sqrt{4x-4}=16-\sqrt{x-1}\)ĐK x lớn hơn hoặc bằng 1
\(6\sqrt{x-1}-3\sqrt{x-1}-2\sqrt{x-1}+\sqrt{x-1}=16\)
\(2\sqrt{x-1}=16\)
\(\sqrt{x-1}=8\)
\(x-1=64\)
\(x=65\)thỏa mãn
Nếu bạn tinh mắt một chút sẽ thấy:
Câu a: \(5\sqrt{2x-1}+2\sqrt{2x-1}-3\sqrt{x}=6\sqrt{2x-1}-2\sqrt{x}\)
Tương đương \(\sqrt{2x-1}=\sqrt{x}\Leftrightarrow\hept{\begin{cases}2x-1=x\\x\ge0\end{cases}}\Leftrightarrow x=1\).
Câu b: \(2\sqrt{x-5}-\sqrt{x-5}=\sqrt{1-x}\).
Tương đương \(\sqrt{x-5}=\sqrt{1-x}\Leftrightarrow\hept{\begin{cases}x\le1\\x-5=1-x\end{cases}}\) (vô nghiệm)
Câu c: \(\sqrt{\left(x+3\right)\left(x-3\right)}-2\sqrt{x-3}=0\)
Tương đương \(\orbr{\begin{cases}x-3=0\\\sqrt{x+3}-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
Ấy chết! Sai ngu ở pt c rồi. Không có nghiệm \(x=1\) nha bạn.
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 \(x=\dfrac{144}{25}\)
Ta có: \(5\sqrt{x-1}-\sqrt{36x-36}+\sqrt{9x-9}=\sqrt{8x+12}\) \(\left(ĐK:x\ge1\right)\)
\(\Leftrightarrow5\sqrt{x-1}-6\sqrt{x-1}+3\sqrt{x-1}=\sqrt{8x+12}\)
\(\Leftrightarrow2\sqrt{x-1}=\sqrt{8x+12}\)
\(\Leftrightarrow\left(2\sqrt{x-1}\right)^2=\left(\sqrt{8x+12}\right)^2\)
\(\Leftrightarrow4.\left(x-1\right)=8x+12\)
\(\Leftrightarrow4x-4=8x+12\)
\(\Leftrightarrow-4x=16\)
\(\Leftrightarrow x=-4\left(L\right)\)
Vậy \(S=\varnothing\)
\(5\sqrt{x-1}-\sqrt{36\left(x-1\right)}+\sqrt{9\left(x-1\right)}=\sqrt{4\left(2x+3\right)}\)
\(5\sqrt{x-1}-6\sqrt{x-1}+3\sqrt{x-1}=2\sqrt{2x+3}\)
\(2\sqrt{x-1}=2\sqrt{2x+3}\)
\(\sqrt{x-1}=\sqrt{2x+3}\)
\(\hept{\begin{cases}2x+3\ge0\\x-1=2x-3\end{cases}}\)
\(\hept{\begin{cases}2x\ge-3\\x-2x=-3+1\end{cases}}\)
\(\hept{\begin{cases}x\ge-\frac{3}{2}\\-x=-2\end{cases}}\)
\(\hept{\begin{cases}x\ge-\frac{3}{2}\\x=2\end{cases}}\)
\(\Rightarrow x=2\)