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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 \(x=\dfrac{144}{25}\)
a/ \(\dfrac{1}{2}.\sqrt{x-1}-\dfrac{3}{2}.\sqrt{9x-9}+24.\sqrt{\dfrac{x-1}{64}}=-17\) ( đkxđ : \(x\ge1\) )
\(\Leftrightarrow\dfrac{1}{2}.\sqrt{x-1}-\dfrac{3}{2}.\sqrt{3^2\left(x-1\right)}+24.\sqrt{\dfrac{x-1}{8^2}}=-17\)
\(\Leftrightarrow\dfrac{1}{2}.\sqrt{x-1}-\dfrac{3.3}{2}.\sqrt{x-1}+\dfrac{24}{8}\sqrt{x-1}=-17\)
\(\Leftrightarrow\) \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\Leftrightarrow\left(\sqrt{x-1}\right)\left(\dfrac{1}{2}-\dfrac{9}{2}+3\right)=-17\)
\(\Leftrightarrow\sqrt{\left(x-1\right)}.\left(-1\right)=-17\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{-17}{-1}=17\)
\(\Leftrightarrow\left(\sqrt{x-1}\right)^2=17^2\)
\(\Leftrightarrow x-1=289\)
\(\Leftrightarrow x=289+1=290\)
vậy x= 290 là nghiệm của phương trình a
b/ \(3x-7\sqrt{x}+4=0\) ( đkxđ : \(x\ge0\) )
\(\Leftrightarrow3x-3\sqrt{x}-4\sqrt{x}+4=0\)
\(\Leftrightarrow3\sqrt{x}\left(\sqrt{x}-1\right)-4\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\left(3\sqrt{x}-4\right)\left(\sqrt{x}-1\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}3\sqrt{x}-4=0\\\sqrt{x}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{4}{3}\\\sqrt{x}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{16}{9}\\x=1\end{matrix}\right.\)
vậy phương trình có tập nghiệm S=\(\left\{1;\dfrac{16}{9}\right\}\)
c/ \(-5x+7\sqrt{x}+12=0\) ( đkxđ: \(x\ge0\) )
\(\Leftrightarrow-\left(5x+5\sqrt{x}-12\sqrt{x}-12\right)=0\)
\(\Leftrightarrow-\left[5\sqrt{x}\left(\sqrt{x}+1\right)-12\left(\sqrt{x}+1\right)\right]\)= 0
\(\Leftrightarrow-\left(5\sqrt{x}-12\right)\left(\sqrt{x}+1\right)=0\)
vì \(x\ge0\Rightarrow\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+1>0\)
\(\Rightarrow5\sqrt{x}-12=0\)
\(\Leftrightarrow\sqrt{x}=\dfrac{12}{5}\Rightarrow x=\dfrac{144}{25}\)
vậy \(x=\dfrac{144}{25}\) là nghiệm của phương trình c
Câu 1:
ĐK: \(x\geq \frac{-3}{2}\)
\(\sqrt{2x+3}=3-\sqrt{5}\)
\(\Rightarrow 2x+3=(3-\sqrt{5})^2=14-6\sqrt{5}\)
\(\Rightarrow x=\frac{11-6\sqrt{5}}{2}\)
Câu 2: ĐK: \(x\geq 0\)
\(\sqrt{5+\sqrt{7x}}=2+\sqrt{7}\)
\(\Rightarrow 5+\sqrt{7x}=(2+\sqrt{7})^2=11+4\sqrt{7}\)
\(\Rightarrow \sqrt{7x}=6+4\sqrt{7}\)
\(\Rightarrow 7x=(6+4\sqrt{7})^2\Rightarrow x=\frac{(6+4\sqrt{7})^2}{7}\)
Câu 3: ĐK: \(x\geq 0\)
\((\sqrt{x}-2)(5-\sqrt{x})=4-x\)
\(\Leftrightarrow 5\sqrt{x}-x-10+2\sqrt{x}=4-x\)
\(\Leftrightarrow 7\sqrt{x}=14\Rightarrow \sqrt{x}=2\Rightarrow x=4\)
Câu 4: ĐK: \(x\ge 1\)
Sửa đề \(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9x-9}+24\sqrt{\frac{x-1}{64}}=-17\)
\(\Leftrightarrow \frac{\sqrt{x-1}}{2}-\frac{3}{2}\sqrt{9}.\sqrt{x-1}+24\sqrt{\frac{1}{64}}\sqrt{x-1}=-17\)
\(\Leftrightarrow \frac{\sqrt{x-1}}{2}-\frac{9\sqrt{x-1}}{2}+3\sqrt{x-1}=-17\)
\(\Leftrightarrow \sqrt{x-1}(\frac{1}{2}-\frac{9}{2}+3)=-17\)
\(\Leftrightarrow -\sqrt{x-1}=-17\Rightarrow \sqrt{x-1}=17\Rightarrow x=17^2+1=290\)
\(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\left(x\text{ ≥}1\right)\)
⇔ \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
⇔ \(-\sqrt{x-1}=-17\)
⇔ \(x=290\left(TM\right)\)
KL..................
a: \(\Leftrightarrow2x+3=14-6\sqrt{5}\)
=>2x=11-6 căn 5
hay \(x=\dfrac{11-6\sqrt{5}}{2}\)
b: \(\Leftrightarrow\sqrt{7x}+5=11+4\sqrt{7}\)
=>căn 7x=6+4 căn 7
=>\(x=\dfrac{\left(6+4\sqrt{7}\right)^2}{7}\)
d: \(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
=>-căn x-1=-17
=>căn x-1=17
=>x-1=289
=>x=290
a: \(\Leftrightarrow\dfrac{2x-3}{x-1}=4\)
=>4x-4=2x-3
=>2x=1
hay x=1/2
b: \(\Leftrightarrow\sqrt{\dfrac{2x-3}{x-1}}=2\)
=>(2x-3)=4x-4
=>4x-4=2x-3
=>2x=1
hay x=1/2(nhận)
c: \(\Leftrightarrow\sqrt{2x+3}\left(\sqrt{2x-3}-2\right)=0\)
=>2x+3=0 hoặc 2x-3=4
=>x=-3/2 hoặc x=7/2
e: \(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
=>căn (x-5)=2
=>x-5=4
hay x=9
a)
ĐKXĐ: \(x> \frac{-5}{7}\)
Ta có: \(\frac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)
\(\Rightarrow 9x-7=\sqrt{7x+5}.\sqrt{7x+5}=7x+5\)
\(\Rightarrow 2x=12\Rightarrow x=6\) (hoàn toàn thỏa mãn)
Vậy......
b) ĐKXĐ: \(x\geq 5\)
\(\sqrt{4x-20}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9x-45}=4\)
\(\Leftrightarrow \sqrt{4}.\sqrt{x-5}+3\sqrt{\frac{1}{9}}.\sqrt{x-5}-\frac{1}{3}\sqrt{9}.\sqrt{x-5}=4\)
\(\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow 2\sqrt{x-5}=4\Rightarrow \sqrt{x-5}=2\Rightarrow x-5=2^2=4\Rightarrow x=9\)
(hoàn toàn thỏa mãn)
Vậy..........
c) ĐK: \(x\in \mathbb{R}\)
Đặt \(\sqrt{6x^2-12x+7}=a(a\geq 0)\Rightarrow 6x^2-12x+7=a^2\)
\(\Rightarrow 6(x^2-2x)=a^2-7\Rightarrow x^2-2x=\frac{a^2-7}{6}\)
Khi đó:
\(2x-x^2+\sqrt{6x^2-12x+7}=0\)
\(\Leftrightarrow \frac{7-a^2}{6}+a=0\)
\(\Leftrightarrow 7-a^2+6a=0\)
\(\Leftrightarrow -a(a+1)+7(a+1)=0\Leftrightarrow (a+1)(7-a)=0\)
\(\Rightarrow \left[\begin{matrix} a=-1\\ a=7\end{matrix}\right.\) \(\Rightarrow a=7\) vì \(a\geq 0\)
\(\Rightarrow 6x^2-12x+7=a^2=49\)
\(\Rightarrow 6x^2-12x-42=0\Leftrightarrow x^2-2x-7=0\)
\(\Leftrightarrow (x-1)^2=8\Rightarrow x=1\pm 2\sqrt{2}\)
(đều thỏa mãn)
Vậy..........
a, ĐKXĐ : \(x\ge1\)
Ta có ; \(PT\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}.\sqrt{9}\sqrt{x-1}+24.\sqrt{\dfrac{1}{64}}\sqrt{x-1}=-17\)
\(\Leftrightarrow\sqrt{x-1}\left(\dfrac{1}{2}-\dfrac{3}{2}\sqrt{9}+24\sqrt{\dfrac{1}{64}}\right)=-17\)
\(\Leftrightarrow-\sqrt{x-1}=-17\)
\(\Leftrightarrow\sqrt{x-1}=17\)
\(\Leftrightarrow x=290\left(TM\right)\)
Vậy ....
b, ĐKXĐ : \(x\ge3\)
Ta có : \(PT\Leftrightarrow x-3-7\sqrt{x-3}+12=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=4\\\sqrt{x-3}=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=16\\x-3=9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=19\\x=12\end{matrix}\right.\) ( TM )
Vậy ..
a) Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\Leftrightarrow-\sqrt{x-1}=-17\)
\(\Leftrightarrow x-1=17^2=289\)
hay x=290
Vậy: S={290}
b) Ta có: \(x-7\sqrt{x-3}+9=0\)
\(\Leftrightarrow x-7\sqrt{x-3}=-9\)
\(\Leftrightarrow x-3-2\cdot\sqrt{x-3}\cdot\dfrac{7}{2}+\dfrac{49}{4}=\dfrac{1}{4}\)
\(\Leftrightarrow\left(\sqrt{x-3}-\dfrac{7}{2}\right)^2=\dfrac{1}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=4\\\sqrt{x-3}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-3=16\\x-3=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=19\\x=12\end{matrix}\right.\)
Vậy: S={19;12}