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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) \(\dfrac{12}{x-1}-\dfrac{8}{x+1}=1\) \(\Leftrightarrow\) \(\dfrac{12\left(x+1\right)-8\left(x-1\right)}{x^2-1}=1\)
\(\Leftrightarrow\) \(\dfrac{12x+12-8x+8}{x^2-1}=1\) \(\Leftrightarrow\) \(\dfrac{4x+20}{x^2-1}=1\)
\(\Leftrightarrow\) \(x^2-1=4x+20\) \(\Leftrightarrow\) \(x^2-4x-21=0\)
giải pt ta có 2 nghiệm : \(x_1=7;x_2=-3\)
vậy phương trình có 2 nghiệm \(x=7;x=-3\)
b) \(\dfrac{16}{x-3}+\dfrac{30}{1-x}=3\) \(\Leftrightarrow\) \(\dfrac{16\left(1-x\right)+30\left(x-3\right)}{\left(x-3\right)\left(1-x\right)}=3\)
\(\Leftrightarrow\) \(\dfrac{16-16x+30x-90}{x-x^2-3+3x}=3\) \(\Leftrightarrow\) \(\dfrac{14x-74}{-x^2+4x-3}=3\)
\(\Leftrightarrow\) \(3\left(-x^2+4x-3\right)=14x-74\)
\(\Leftrightarrow\) \(-3x^2+12x-9=14x-74\)
\(\Leftrightarrow\) \(3x^2-2x-65=0\)
giải pt ta có 2 nghiệm : \(x_1=5;x_2=\dfrac{-13}{3}\)
vậy phương trình có 2 nghiệm \(x=5;x=\dfrac{-13}{3}\)
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
15
\(\dfrac{7}{x-2}\)+\(\dfrac{8}{x-5}\)=3 (x khác 2 khác 5)
\(\Leftrightarrow\)7*(x-5)+8(x-2)=3(x-2)(x-5)
\(\Leftrightarrow\)15x-51=3x^2-21x+30\(\Leftrightarrow\)3x^2-36x+81=0
\(\Leftrightarrow\)\(\begin{matrix}&\end{matrix}\)\(\left[{}\begin{matrix}9\\3\end{matrix}\right.\) tmđk
16\(\dfrac{x^2-3x+6}{x^2-9}\)=\(\dfrac{1}{x-3}\)(x khác +_3)
\(\Leftrightarrow\)x^2-3x+6=x+3
\(\Leftrightarrow\)x^2-4x+3=0\(\Leftrightarrow\)\(\left[{}\begin{matrix}3loại\\1\end{matrix}\right.\)
vậy x=1 là nghiệm của pt
17 \(\dfrac{3}{x^2-4}\) = \(\dfrac{1}{x-2}+\dfrac{1}{x+2}\)
<=> x + 2 + x - 2 = 3
<=> 2x = 3
<=> x = \(\dfrac{3}{2}\)
a, \(\sqrt{9x+9}-4\sqrt{\dfrac{x+1}{4}}=5\) \(x\ge-1\)
\(\Leftrightarrow3\sqrt{x+1}-2\sqrt{x+1}=5\)
\(\Leftrightarrow x+1=25\Leftrightarrow x=24\)
2) "biểu thức"=\(\sqrt{x-5}-4\sqrt{x-5}-\sqrt{x-5}=12\Leftrightarrow4\sqrt{x-5}=12\Leftrightarrow\sqrt{x-5}=3\Leftrightarrow x=14\)
Kl: x=14
3) "biểu thức"=\(4\sqrt{x-1}-3\sqrt{x-1}+\sqrt{x-1}=5\Leftrightarrow2\sqrt{x-1}=5\Leftrightarrow\sqrt{x-1}=\dfrac{5}{2}\Leftrightarrow x=\left(\dfrac{5}{2}\right)^2+1=\dfrac{29}{4}\)
Kl: x=29/4
\(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\left(x\text{ ≥}1\right)\)
⇔ \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
⇔ \(-\sqrt{x-1}=-17\)
⇔ \(x=290\left(TM\right)\)
KL..................
a) \(\frac{1}{x-1+\sqrt{x^2-2x+3}}+\frac{1}{x-1-\sqrt{x^2-2x+3}}=1\)
ĐKXĐ : \(x\inℝ\)
\(\Leftrightarrow\frac{x-1-\sqrt{x^2-2x+3}}{\left(x-1+\sqrt{x^2-2x+3}\right)\left(x-1-\sqrt{x^2-2x+3}\right)}+\frac{x-1+\sqrt{x^2-2x+3}}{\left(x-1+\sqrt{x^2-2x+3}\right)\left(x-1-\sqrt{x^2-2x+3}\right)}=\frac{\left(x-1+\sqrt{x^2-2x+3}\right)\left(x-1-\sqrt{x^2-2x+3}\right)}{\left(x-1+\sqrt{x^2-2x+3}\right)\left(x-1-\sqrt{x^2-2x+3}\right)}\)
\(\Rightarrow2x-2=\left[\left(x-1\right)+\left(\sqrt{x^2-2x+3}\right)\right]\left[\left(x-1\right)-\left(\sqrt{x^2-2x+3}\right)\right]\)
\(\Leftrightarrow2x-2=\left(x-1\right)^2-\left(\sqrt{x^2-2x+3}\right)^2\)
\(\Leftrightarrow2x-2=x^2-2x+1-\left(x^2-2x+3\right)\)
\(\Leftrightarrow2x-2=x^2-2x+1-x^2+2x-3\)
\(\Leftrightarrow2x-2=-2\)
\(\Leftrightarrow2x=0\)
\(\Leftrightarrow x=0\)
Vậy phương trình có nghiệm duy nhất x = 0