Câu 3 (2 điểm). Tìm xx biết:
1) √4(1−3x)+√9(1−3x)=104(1−3x)+9(1−3x)=10
2) (√x+1)(2√x−3)−2x=−4(x+1)(2x−3)−2x=−4
3) √2x+1−x+1=0
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\(A=\left(12\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\)
\(A=\left(24\sqrt{7}-2\sqrt{3}-\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\)
\(A=168-2\sqrt{21}-7+2\sqrt{21}\)
\(A=161\)
Bài làm :
Bình phương cả 2 vế lên ; ta được :
\(4\sqrt{3}+\sqrt{x}=4+4\sqrt{3}+3\)
\(\Leftrightarrow4\sqrt{3}+\sqrt{x}=4\sqrt{3}+7\)
\(\Leftrightarrow\sqrt{x}=7\)
\(\Leftrightarrow x=49\)
Vậy x=49
đk: \(x\ge0\)
Ta có: \(\sqrt{4\sqrt{3}+\sqrt{x}}=2+\sqrt{3}\)
\(\Leftrightarrow\left(\sqrt{4\sqrt{3}+\sqrt{x}}\right)^2=\left(2+\sqrt{3}\right)^2\)
\(\Leftrightarrow4\sqrt{3}+\sqrt{x}=7+4\sqrt{3}\)
\(\Leftrightarrow\sqrt{x}=7\)
\(\Rightarrow x=49\left(tm\right)\)
Vậy x = 49
ĐKXĐ : x ≠ -1
pt ⇔ \(\sqrt{3^2\left(x+1\right)}+\sqrt{x+1}=20\)
⇔ \(3\sqrt{x+1}+\sqrt{x+1}=20\)
⇔ \(4\sqrt{x+1}=20\)
⇔ \(\sqrt{x+1}=5\)
⇔ \(x+1=25\)
⇔ \(x=24\)( tm )
\(ĐKXĐ:x\ge-1\)
\(\sqrt{9x+9}+\sqrt{x+1}=20\)
\(\Leftrightarrow\sqrt{9\left(x+1\right)}+\sqrt{x+1}=20\)
\(\Leftrightarrow\sqrt{9}.\sqrt{x+1}+\sqrt{x+1}=20\)
\(\Leftrightarrow3\sqrt{x+1}+\sqrt{x+1}=20\)
\(\Leftrightarrow4\sqrt{x+1}=20\)
\(\Leftrightarrow\sqrt{x+1}=5\)
\(\Leftrightarrow x+1=25\)
\(\Leftrightarrow x=24\)( thỏa mãn ĐKXĐ )
Vậy \(x=24\)
a) \(\Delta ABC\)vuông tại A , đường cao AH
Ta có các hệ thức sau
\(HB.HC=AH^2\)
\(AB^2=BH.BC\)
\(AC=CH.BC\)
\(\frac{1}{AH^2}=\frac{1}{AB^2}+\frac{1}{AC^2}\)
Ta có: \(\frac{1}{BH.BC}+\frac{1}{CH.BC}=\frac{1}{AB^2}+\frac{1}{AC^2}=\frac{1}{AH^2}\)
mà \(\frac{1}{HB.HC}=\frac{1}{AH^2}\)
\(\Rightarrow\frac{1}{HB.HC}=\frac{1}{BH.BC}+\frac{1}{CH.BC}\)( đpcm )
\(A=\left(\frac{1}{x+2\sqrt{x}}-\frac{1}{\sqrt{x}+2}\right)\div\frac{1-\sqrt{x}}{x+4\sqrt{x}+4}\)
ĐKXĐ : \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)
\(=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\right)\div\frac{1-\sqrt{x}}{\left(\sqrt{x}+2\right)^2}\)
\(=\frac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\times\frac{\left(\sqrt{x}+2\right)^2}{1-\sqrt{x}}\)
\(=\frac{\sqrt{x}+2}{\sqrt{x}}\)
Để A = 5/3
=> \(\frac{\sqrt{x}+2}{\sqrt{x}}=\frac{5}{3}\)( \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\))
=> \(3\left(\sqrt{x}+2\right)=5\sqrt{x}\)
=> \(3\sqrt{x}+6=5\sqrt{x}\)
=> \(5\sqrt{x}-3\sqrt{x}=6\)
=> \(2\sqrt{x}=6\)
=> \(\sqrt{x}=3\)
=> \(x=9\)( tm )