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a)\(\sqrt{3x+1}+2x=\sqrt{x-4}-5\left(ĐKXĐ:x\ge4\right)\)
\(\Leftrightarrow\left(\sqrt{3x+1}-\sqrt{x-4}\right)+\left(2x+5\right)=0\)
\(\Leftrightarrow\frac{3x+1-x+4}{\sqrt{3x+1}+\sqrt{x-4}}+\left(2x+5\right)=0\)
\(\Leftrightarrow\frac{2x+5}{\sqrt{3x+1}+\sqrt{x-4}}+\left(2x+5\right)=0\)
\(\Leftrightarrow\left(2x+5\right)\left(\frac{1}{\sqrt{3x+1}+\sqrt{x-4}}+1\right)=0\)
a') (tiếp)
\(\Leftrightarrow\orbr{\begin{cases}2x+5=0\\\frac{1}{\sqrt{3x+1}+\sqrt{x-4}}+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2,5\left(KTMĐKXĐ\right)\\\frac{1}{\sqrt{3x+1}+\sqrt{x-4}}+1=0\end{cases}}\)
Xét phương trình \(\frac{1}{\sqrt{3x+1}+\sqrt{x-4}}+1=0\)(1)
Với mọi \(x\ge4\), ta có:
\(\sqrt{3x+1}>0\); \(\sqrt{x-4}\ge0\)
\(\Rightarrow\sqrt{3x+1}+\sqrt{x-4}>0\Rightarrow\frac{1}{\sqrt{3x+1}+\sqrt{x-4}}>0\)
\(\Rightarrow\frac{1}{\sqrt{3x+1}+\sqrt{x-4}}+1>0\)
Do đó phương trình (1) vô nghiệm.
Vậy phương trình đã cho vô nghiệm.
a)\(\sqrt{4x+20}\) +\(\sqrt{x-5}\) -\(\dfrac{1}{3}\)\(\sqrt{9x-45}\)=4 ; ĐKXĐ : x ≥_+ 5
⇔ \(\sqrt{2^2x+2^2.5}\) +\(\sqrt{x-5}\) -\(\dfrac{1}{3}\)\(\sqrt{3^2x-3^2.5}\) =4
⇔ 2\(\sqrt{x+5}\) +\(\sqrt{x-5}\) -\(\dfrac{1}{3}\)3\(\sqrt{x-5}\) =4 ⇔ 2\(\sqrt{x+5}\) +\(\sqrt{x-5}\) -\(\sqrt{x-5}\) =4⇔2\(\sqrt{x+5}\)=4(tm)
⇔\(\sqrt{x+5}\)=2⇔x+5=4 ⇔x=-1
Vậy x=-1
b) \(\sqrt{x^2-36}\) - \(\sqrt{x-6}\) =0 ; ĐKXĐ: x≥_+6
⇔ \(\sqrt{\left(x-6\right)\left(x+6\right)}\) - \(\sqrt{x-6}\) =0 ⇔ \(\sqrt{x-6}\).\(\sqrt{x+6}\) - \(\sqrt{x-6}\) =0
⇔ \(\sqrt{x-6}\)(\(\sqrt{x+6}\) -1 )=0 ⇔\([\) \(\begin{matrix}\sqrt{x-6}&=0\\\sqrt{x+6}-1&=0\end{matrix}\) ⇔ \([\) \(\begin{matrix}x-6&=0\\x+6-1&=0\end{matrix}\) ⇔\([\) \(\begin{matrix}x&=6\left(ktm\right)\\x&=-5\left(tm\right)\end{matrix}\)
Vậy x=-5
c) \(\sqrt{4-x^2}\) -x +2 =0 ; ĐKXĐ: -2≤x≤2
⇔ \(\sqrt{\left(2-x\right)\left(2+x\right)}\) -x+2 =0 ⇔ \(\sqrt{\left(2-x\right)\left(2+x\right)}\) -(x-2)=0
⇔ \(\sqrt{\left(2-x\right)\left(2+x\right)}\) =(x-2) ⇔ (2-x)(2+x)=(x-2)2 ⇔ 4-x2 = x2-4x+4 ⇔ -x2-x2+4x=4-4
⇔-2x2+4x=0 ⇔ -2x(x-2)=0 ⇔ \([\) \(\begin{matrix}-2x&=0\\x-2&=0\end{matrix}\) ⇔\([\) \(\begin{matrix}x&=0\left(tm\right)\\x&=2\left(tm\right)\end{matrix}\)
Vậy S=\(\left\{0;2\right\}\)
d) \(\sqrt{\left(2x-3\right)\left(x-1\right)}-\sqrt{x-1}=0\) ; ĐKXĐ: x≥\(\dfrac{3}{2}\);x ≥ 1
⇔\(\sqrt{2x-3}.\sqrt{x-1}-\sqrt{x-1}=0\) ⇔ \(\sqrt{x-1}.\left(\sqrt{2x-3}-1\right)=0\)
⇔ \(\left[{}\begin{matrix}\sqrt{x-1}=0\\\sqrt{2x-3}-1=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x-1=0\\2x-3-1=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=1\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)
Vậy s=\(\left\{1:2\right\}\)
ĐKXĐ: \(x\ne-1\)
\(\frac{1}{\sqrt{x^2+3}}+\frac{1}{\sqrt{1+3x^2}}=\frac{2}{x+1}\)
\(\Leftrightarrow\frac{x+1}{\sqrt{x^2+3}}-1+\frac{x+1}{\sqrt{3x^2+1}}-1=0\)
\(\Leftrightarrow\frac{x+1-\sqrt{x^2+3}}{\sqrt{x^2+3}}+\frac{x+1-\sqrt{3x^2+1}}{\sqrt{3x^2+1}}=0\)
\(\Leftrightarrow\frac{x^2+2x+1-x^2-3}{\sqrt{x^2+3}\left(x+1+\sqrt{x^2+3}\right)}+\frac{x^2+2x+1-3x^2-1}{\sqrt{3x^2+1}\left(x+1+\sqrt{3x^2+1}\right)}=0\)
\(\Leftrightarrow\frac{2\left(x-1\right)}{\sqrt{x^2+3}\left(x+1+\sqrt{x^2+3}\right)}+\frac{-2x\left(x-1\right)}{\sqrt{3x^2+1}\left(x+1+\sqrt{3x^2+1}\right)}=0\)
\(\Leftrightarrow2\left(x-1\right)\left(\frac{1}{\sqrt{x^2+3}\left(x+1+\sqrt{x^2+3}\right)}-\frac{1}{\sqrt{\frac{1}{x^2}+3}\left(\frac{1}{x}+1+\sqrt{\frac{1}{x^2}+3}\right)}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\\sqrt{x^2+3}\left(x+1+\sqrt{x^2+3}\right)=\sqrt{\frac{1}{x^2}+3}\left(\frac{1}{x}+1+\sqrt{\frac{1}{x^2}+3}\right)\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x^2=\frac{1}{x^2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\left(tmđkxđ\right)\\x=-1\left(ktmđkxđ\right)\end{cases}\Rightarrow}x=1}\)
Vậy nghiệm của pt trên là x=1
Xét tử:
\(2\sqrt{1-3x}+\sqrt[3]{x+9}-2=2\left(\sqrt{1-3x}+\frac{3x-5}{4}\right)+\left(\sqrt[3]{x+9}-\frac{-3x+1}{2}\right)\)
\(=2.\frac{1-3x-\frac{9x+25-30x}{16}}{\sqrt{1-3x}-\frac{3x-5}{4}}+\frac{x+9-\left(\frac{-3x+1}{2}\right)^3}{\sqrt[3]{\left(x+9\right)^2}+\sqrt[3]{x+9}.\frac{-3x+1}{2}+\left(\frac{-3x+1}{2}\right)^2}\)
\(=\frac{-18\left(x+1\right)^2}{\sqrt{1-3x}-\frac{3x-5}{4}}+\frac{\frac{\left(x+1\right)\left(27x^2-54x+71\right)}{8}}{\sqrt[3]{\left(x+9\right)^2}+\sqrt[3]{x+9}.\frac{-3x+1}{2}+\left(\frac{-3x+1}{2}\right)^2}\)
Xét mẫu : x2-2x-3=(x+1)(x-3)
\(\Rightarrow A=\frac{\frac{-18\left(x+1\right)}{\sqrt{1-3x}-\frac{3x-5}{4}}+\frac{\frac{27x^2-54x+71}{8}}{\sqrt[3]{\left(x+9\right)^2}+\sqrt[3]{\left(x+9\right)}.\frac{-3x+1}{2}+\left(\frac{-3x+1}{2}\right)^2}}{x-3}\)
\(lim_{x\rightarrow-1}A=\frac{19}{48}\)
Gõ nhờ tý nhé, ko phải đáp án đâu
Bài 1:
a, \(\sqrt{2x+5}=\sqrt{1-x}\)
\(\Rightarrow2x+5=1-x\Rightarrow2x+x=1-5\)
\(\Rightarrow3x=-4\Rightarrow x=-\dfrac{4}{3}\)
b, \(\sqrt{x^2-x}=\sqrt{3-x}\)
\(\Rightarrow x^2-x=3-x\)
\(\Rightarrow x^2-x+x=3\Rightarrow x^2=3\)
\(\Rightarrow x=\pm\sqrt{3}\)
c, \(\sqrt{2x^2-3}=\sqrt{4x-3}\)
\(\Rightarrow2x^2-3=4x-3\)
\(\Rightarrow2x^2-4x=0\Rightarrow2x.\left(x-2\right)=0\)
\(\Rightarrow x.\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Chúc bạn học tốt!!!
ngu thế bạn :) bạn chỉ cần ghi
\(dkxd\Leftrightarrow x\ne\frac{2}{3}\)
là xong có j đâu :))
B cứ giải bình thường đi
Đặt \(\hept{\begin{cases}\sqrt{4x+1}=a\\\sqrt{3x-2}=b\end{cases}}\)
Rồi làm tiếp