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a,\(\sqrt{3x+1}=3x-1\) Đk:\(x\ge\dfrac{-1}{3}\)
\(< =>3x+1=9x^2-6x+1\)
\(< =>9x-9x^2=0\)
\(< =>9x\left(1-x\right)=0\)
\(< =>x=0\) hoặc \(x=1\)
b,\(2+\sqrt{3x-5}=x+1\) Đk:\(x\ge\dfrac{5}{3}\)
\(< =>\sqrt{3x-5}=x-1\)
\(< =>3x-5=x^2-2x+1\)
\(< =>x^2+x+6=0\)(vô lý vì \(x^2\ge\dfrac{25}{9},x\ge\dfrac{5}{3}\))
=>\(x\in\varnothing\)
c,Đk : \(x\ge\dfrac{-7}{5}\)
\(\)\(\dfrac{5x+7}{x+3}=16\)
\(< =>5x+7=16x+48\)
\(< =>-11x=41 \)
\(< =>x=\dfrac{-41}{11}\)(ko tm đk)
\(=>x\in\varnothing\)
d,tương tự câu c bình phương 2 vế cũng ra \(x\in\varnothing\)
a,\(1+\sqrt{3x+1}=3x\)(ĐK:\(x>-\frac{1}{3}\))
\(\Leftrightarrow\sqrt{3x+1}=3x-1\)
\(\Leftrightarrow3x+1=9x^2-6x+1\)
\(\Leftrightarrow9x^2-9x=0\)
\(\Leftrightarrow9x\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\left(tm\right)\\x=1\left(tm\right)\end{cases}}\)
b,\(\sqrt{2+\sqrt{3x-5}}=\sqrt{x+1}\)(ĐK:\(x>-\frac{5}{3}\))
\(\Leftrightarrow2+\sqrt{3x-5}=x+1\)
\(\Leftrightarrow2+3x-5+2.2\sqrt{3x-5}=x+1\)
\(\Leftrightarrow3x-3-x-1=4\sqrt{3x-5}\)
\(\Leftrightarrow2x-4=4\sqrt{3x-5}\)
\(\Leftrightarrow4x^2-16x+16=48x-80\)
\(\Leftrightarrow4x^2-64x-64=0\)
\(\Delta=64^2-4.\left(-64\right)=4352\)
\(\orbr{\begin{cases}x_1=\frac{64-\sqrt{4352}}{8}=8-2\sqrt{17}\left(tm\right)\\x_2=\frac{64+\sqrt{4352}}{8}=8+2\sqrt{17}\left(tm\right)\end{cases}}\)
c,Cho biểu thức trong căn nhận giá trị 16 mà giải
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) \(\text{Đ}K\text{X}\text{Đ}:\frac{3}{2}\le x\le\frac{5}{2}\)
Áp dụng BĐT Bunhiacopxki ta có:
\(VT=\sqrt{2x-3}+\sqrt{5-2x}\le\sqrt{2\left(2x-3+5-2x\right)}=2\)
Dấu '=' xảy ra khi \(\sqrt{2x-3}=\sqrt{5-2x}\Leftrightarrow x=2\)
Lại có: \(VP=3x^2-12x+14=3\left(x-2\right)^2+2\ge2\)
Dấu '=' xảy ra khi x=2
Do đó VT=VP khi x=2
b) ĐK: \(x\ge0\). Ta thấy x=0 k pk là nghiệm của pt, chia 2 vế cho x ta có:
\(x^2-2x-x\sqrt{x}-2\sqrt{x}+4=0\Leftrightarrow x-2-\sqrt{x}-\frac{2}{\sqrt{x}}+\frac{4}{x}=0\)
\(\Leftrightarrow\left(x+\frac{4}{x}\right)-\left(\sqrt{x}+\frac{2}{\sqrt{x}}\right)-2=0\)
Đặt \(\sqrt{x}+\frac{2}{\sqrt{x}}=t>0\Leftrightarrow t^2=x+4+\frac{4}{x}\Leftrightarrow x+\frac{4}{x}=t^2-4\), thay vào ta có:
\(\left(t^2-4\right)-t-2=0\Leftrightarrow t^2-t-6=0\Leftrightarrow\left(t-3\right)\left(t+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}t=3\\t=-2\end{cases}}\)
Đối chiếu ĐK của t
\(\Rightarrow t=3\Leftrightarrow\sqrt{x}+\frac{2}{\sqrt{x}}=3\Leftrightarrow x-3\sqrt{x}+2=0\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=1\end{cases}}\)
a) ĐKXĐ: \(\left\{{}\begin{matrix}5-x\ge0\\x-3\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-x\ge-5\\x\ge3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le5\\x\ge3\end{matrix}\right.\Leftrightarrow3\le x\le5\)
Ta có: \(\sqrt{5-x}+\sqrt{x-3}=\sqrt{2}\)
\(\Leftrightarrow\left(\sqrt{5-x}+\sqrt{x-3}\right)^2=\left(\sqrt{2}\right)^2\)
\(\Leftrightarrow5-x+2\cdot\sqrt{\left(5-x\right)\cdot\left(x-3\right)}+x-3=2\)
\(\Leftrightarrow2+2\cdot\sqrt{\left(5-x\right)\cdot\left(x-3\right)}=2\)
\(\Leftrightarrow2\cdot\sqrt{\left(5-x\right)\cdot\left(x-3\right)}=0\)
mà \(2\ne0\)
nên \(\sqrt{\left(5-x\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\left(5-x\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5-x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\left(nhận\right)\\x=3\left(nhận\right)\end{matrix}\right.\)
Vậy: S={3;5}
b) ĐKXĐ: \(\left\{{}\begin{matrix}x^2-4\ge0\\x-2\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)\left(x+2\right)\ge0\\x-2\ge0\end{matrix}\right.\Leftrightarrow x-2\ge0\)\(\Leftrightarrow x\ge2\)
Ta có: \(\sqrt{x^2-4}=2\sqrt{x-2}\)
\(\Leftrightarrow\sqrt{x-2}\cdot\sqrt{x+2}-2\cdot\sqrt{x-2}=0\)
\(\Leftrightarrow\sqrt{x-2}\cdot\left(\sqrt{x+2}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\\sqrt{x+2}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x+2=4\end{matrix}\right.\Leftrightarrow x=2\)
Vậy: S={2}
a: =>\(\sqrt{3x-5}+2=x+1\)
\(\Leftrightarrow\sqrt{3x-5}=x-1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=1\\x^2-2x+1-3x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
b: \(\Leftrightarrow x-15\sqrt{x}+56=x+11\)
=>-15 căn x=-45
=>x=9
c: =>căn 3x+1=3x-1
\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{1}{3}\\9x^2-6x+1-3x-1=0\end{matrix}\right.\Leftrightarrow x=1\)
d: =>(3x+7)/(x+3)=16
=>16x+48=3x+7
=>13x=-41
=>x=-41/13
@Nguyễn Huy Thắng@Mysterious Person@bảo nam trần@Lightning Farron@Thiên Thảo@Sky SơnTùng
a ) \(1+\sqrt{3x+1}=3x\) ( ĐKXĐ : \(x\ge-\dfrac{1}{3}\) )
\(\Leftrightarrow\sqrt{3x+1}=3x-1\)
\(\Leftrightarrow3x+1=\left(3x-1\right)^2\)
\(\Leftrightarrow3x+1-\left(3x-1\right)^2=0\)
\(\Leftrightarrow3x+1-9x^2+6x-1=0\)
\(\Leftrightarrow9x^2-9x=0\)
\(\Leftrightarrow9x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}9x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\)
Vậy phương trình có nghiệm x = 0 hoặc x = 1 .