giải phương trình:
\(\dfrac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)
giải phương trình
a, \(\sqrt{4x-20}+3\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
b, \(2x-x^2+\sqrt{6x^2-12x+7}=0\)
c, \(\dfrac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)
Giair các phương trình sau:
a) \(\dfrac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)
\(ĐK:x>-\dfrac{5}{7}\\ PT\Leftrightarrow7x+5=9x-7\Leftrightarrow x=6\left(tm\right)\)
giải các phương trình sau:
a. \(2\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}=28\)
b. \(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\)
c. \(\sqrt{\dfrac{3x-2}{x+1}}=3\)
Lời giải:
a. ĐKXĐ: $x\geq 0$
$2\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}=28$
$\Leftrightarrow 2\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=28$
$\Leftrightarrow 13\sqrt{2x}=28$
$\Leftrightarrow \sqrt{2x}=\frac{28}{13}$
$\Leftrightarrow 2x=\frac{784}{169}$
$\Leftrightarrow x=\frac{392}{169}$
b. ĐKXĐ: $x\geq 5$
PT $\Leftrightarrow \sqrt{4}.\sqrt{x-5}+\sqrt{x-5}-\frac{1}{3}.\sqrt{9}.\sqrt{x-5}=4$
$\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4$
$\Leftrightarrow 2\sqrt{x-5}=4$
$\Leftrightarrow \sqrt{x-5}=2$
$\Leftrightarrow x-5=4$
$\Leftrightarrow x=9$ (tm)
c. ĐKXĐ: $x\geq \frac{2}{3}$ hoặc $x< -1$
PT $\Leftrightarrow \frac{3x-2}{x+1}=9$
$\Rightarrow 3x-2=9(x+1)$
$\Leftrightarrow x=\frac{-11}{6}$ (tm)
Giải các pt sau: a)\(\dfrac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\) b)\(\sqrt{4x-20}+3\sqrt{\dfrac{x+5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
a) \(\dfrac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\) (1)
\(\Leftrightarrow9x-7=\sqrt{\left(7x+5\right)\left(7x+5\right)}\)
\(\Leftrightarrow9x-\sqrt{\left(7x+5\right)\left(7x+5\right)}=7\)
\(\Leftrightarrow9x-\sqrt{\left(7x+5\right)^2}=7\)
\(\Leftrightarrow9x-\left|7x+5\right|=7\)
\(\Leftrightarrow\left[{}\begin{matrix}9x-\left(7x+5\right)=7\left(đk:7x+5\ge0\right)\\9x-\left[-\left(7x+5\right)\right]=7\left(đk:7x+5< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\left(đk:x\ge-\dfrac{5}{7}\right)\\x=\dfrac{1}{8}\left(đk:x< -\dfrac{5}{7}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x\in\varnothing\end{matrix}\right.\)
\(\Leftrightarrow x=6\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{6\right\}\)
b) \(\sqrt{4x-20}+3\sqrt{\dfrac{x+5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\) (2)
\(\Leftrightarrow\sqrt{4\left(x-5\right)}+3\cdot\dfrac{\sqrt{x+5}}{3}-\dfrac{1}{3}\cdot\sqrt{9\left(x-5\right)}=4\)
\(\Leftrightarrow\sqrt{4}\sqrt{x-5}+\sqrt{x+5}-\dfrac{1}{3}\cdot\sqrt{9}\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x+5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x+5}-\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}+\sqrt{x+5}=4\)
\(\Leftrightarrow\sqrt{x-5}=4-\sqrt{x+5}\)
\(\Leftrightarrow x-5=\left(4-\sqrt{x+5}\right)^2\)
\(\Leftrightarrow x-5=16-8\sqrt{x+5}+x+5\)
\(\Leftrightarrow-5=16-8\sqrt{x+5}+5\)
\(\Leftrightarrow-5=21-8\sqrt{x+5}\)
\(\Leftrightarrow8\sqrt{x+5}=21+5\)
\(\Leftrightarrow8\sqrt{x+5}=26\)
\(\Leftrightarrow\sqrt{x+5}=\dfrac{13}{4}\)
\(\Leftrightarrow x+5=\dfrac{169}{16}\)
\(\Leftrightarrow x=\dfrac{169}{16}-5\)
\(\Leftrightarrow x=\dfrac{89}{16}\)
Vậy tập nghiệm phương trình (2) là \(S=\left\{\dfrac{89}{16}\right\}\)
Giải phương trình \(x^2-7x=6\sqrt{x+5}-30\)
ĐKXĐ: \(x\ge-5\)
\(\Leftrightarrow x^2-8x+16+x+5-6\sqrt{x+5}+9=0\)
\(\Leftrightarrow\left(x-4\right)^2+\left(\sqrt{x+5}-3\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-4=0\\\sqrt{x+5}-3=0\end{matrix}\right.\)
\(\Leftrightarrow x=4\)
1, \(K=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)
2, \(\sqrt{x-3}-2.\sqrt{x^2-3x}=0\)
3, \(\dfrac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)
4, \(x-5\sqrt{x}+4=0\)
1,\(K=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{x}}\)
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}\right)\)\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{5}-1\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}\right)\)
\(=\dfrac{1}{\sqrt{2}}\left(\left|\sqrt{5}-1\right|+\sqrt{5}+1\right)\)\(=\dfrac{1}{\sqrt{2}}\left|\sqrt{5}-1+\sqrt{5}+1\right|=\dfrac{1}{\sqrt{2}}.2\sqrt{5}\)\(=\sqrt{10}\)
2, \(\sqrt{x-3}-2\sqrt{x^2-3x}=0\left(đk:x\ge3\right)\)
\(\Leftrightarrow\sqrt{x-3}\left(1-2\sqrt{x}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\1-2\sqrt{x}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=\left(\dfrac{1}{2}\right)^2=\dfrac{1}{4}\left(ktm\right)\end{matrix}\right.\)
Vậy pt có nghiệm x=3
3, \(\dfrac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\left(đk:x>-\dfrac{5}{7}\right)\)
\(\Leftrightarrow9x-7=7x+5\)
\(\Leftrightarrow x=6\left(tm\right)\)
4, \(x-5\sqrt{x}+4=0\)(đk: \(x\ge0\))
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}=4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=16\end{matrix}\right.\) (tm)
Vậy...
1) Bạn tự làm
2) ĐK: \(x\ge3\)
PT \(\Leftrightarrow\sqrt{x-3}\left(1-2\sqrt{x}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\2\sqrt{x}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{4}\left(loại\right)\end{matrix}\right.\)
Vậy ...
3) ĐK: \(x>-\dfrac{5}{7}\)
PT \(\Rightarrow9x-7=7x+5\) \(\Leftrightarrow x=6\)
Vậy ...
4) ĐK: \(x\ge0\)
PT \(\Leftrightarrow x-4\sqrt{x}-\sqrt{x}+4=0\)
\(\Leftrightarrow\left(\sqrt{x}-4\right)\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=4\\\sqrt{x}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=16\\x=1\end{matrix}\right.\)
Vậy ...
Giải phương trình :
\(\left(\sqrt{x+5}-\sqrt{x-2}\right)\left(1+\sqrt{x^2+7x+10}\right)=3\)
Tìm x thõa mãn: \(\dfrac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)
Sắp off rồi nên làm nhanh :))
ĐKXĐ: \(x>-\dfrac{5}{7}\)
\(\dfrac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)
\(\Leftrightarrow9x-7=7x+5\)
\(\Leftrightarrow x=6\) ( thỏa mãn đk )
Vậy x = 6
Akai HarumaHung nguyenNgô Thanh SangAce Legona
Tìm điều kiện xác định quy đồng cái ra luôn
Giải phương trình:
\(\frac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)
\(\frac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)
\(\Leftrightarrow9x-7=\sqrt{7x+5}.\sqrt{7x+5}\)
\(\Leftrightarrow9x-7=7x+5\)
\(\Leftrightarrow9x-7x=7+5\)
\(\Leftrightarrow2x=12\)
\(\Leftrightarrow x=6\)
Vậy nghiệm phương trình là x = 6.
Điều kiện: \(7x+5\ge0\Leftrightarrow x\ge-\frac{5}{7}\)
Kết hợp điều kiện, x = 6 là nghiệm của phương trình.