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\(a,\sqrt{x^4}=7\Leftrightarrow x^2=7\Leftrightarrow x=\pm\sqrt{7}\)
\(Dk:x\ge\frac{2}{3};\sqrt{3x-2}=4\Leftrightarrow3x-2=16\Leftrightarrow3x=18\Leftrightarrow x=6\left(tm\right)\)
\(dk:x\ge\frac{3}{2};\sqrt{2x-3}=\sqrt{x-1}\Leftrightarrow2x-3=x-1\Leftrightarrow x=2\left(tm\right)\)
\(dk:x\ge0;x-10\sqrt{x}+25=0\Leftrightarrow\left(\sqrt{x}-5\right)^2=0\Leftrightarrow\sqrt{x}=5\Leftrightarrow x=25\left(tm\right)\)
\(\sqrt{2x}< 3\Leftrightarrow\sqrt{2}.\sqrt{x}< 3\Leftrightarrow0\le\sqrt{x}< \sqrt{4,5}\Leftrightarrow0\le x< 4,5\)
\(h,dk:x\ge3;\sqrt{\left(x-1\right)^2}=3x-9\Leftrightarrow\left|x-1\right|=3x-9\Leftrightarrow x-1=3x-9\left(x\ge3\right)\Leftrightarrow x=4\left(tm\right)\)
a)\(\sqrt{3x+2}=2-\sqrt{3}\)
\(\Leftrightarrow3x+2=\left(2-\sqrt{3}\right)^2\)
\(\Leftrightarrow3x+2=7-4\sqrt{3}\)
\(\Leftrightarrow3x=7-2-4\sqrt{3}\)
\(\Leftrightarrow3x=5-4\sqrt{3}\)
\(\Leftrightarrow x=\dfrac{5}{3}-\dfrac{4\sqrt{3}}{3}\)
\(\Leftrightarrow x=\dfrac{5-4\sqrt{3}}{3}\)
b) \(\sqrt{x^2-4x+4}=49\)
\(\Leftrightarrow\sqrt{\left(x-2\right)^2}=49\)
\(\Leftrightarrow\left|x-2\right|=49\)\
\(\Leftrightarrow\left[{}\begin{matrix}x-2=49\\-x+2=49\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=51\\x=-47\end{matrix}\right.\)
c) \(\sqrt{x+1}=x-1\)
ĐKXĐ: \(x-1\ge0\Rightarrow x\ge1\)
\(\Leftrightarrow x+1=\left(x-1\right)^2\)
\(\Leftrightarrow x+1=x^2-2x+1\)
\(\Leftrightarrow-x^2+2x+x=-1+1\)
\(\Leftrightarrow3x-x^2=0\)
\(\Leftrightarrow x\left(3-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\3-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(lo\text{ại}\right)\\x=3\left(nh\text{ậ}n\right)\end{matrix}\right.\)
d)e) lát mình làm sau
2,\(pt\Leftrightarrow12\left(\sqrt{x+1}-2\right)+x^2+x-12=0\)
\(\Leftrightarrow12\cdot\frac{x-3}{\sqrt{x+1}+2}+\left(x-3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)=0\)
Vì \(\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)\ge0\left(\forall x>-1\right)\)
\(\Rightarrow x=3\)
a) Đk: \(-\dfrac{1}{3}\le x\le2\)
\(\sqrt{3x+1}+\sqrt{2-x}=1\Leftrightarrow\sqrt{-3x^2+5x+2}=-x-1\)
Ta có: \(VT\ge0\) ; \(VP< 0\forall-\dfrac{1}{3}\le x\le2\)
Kl: ptvn
b) \(x^2+5x+9=\left(x+5\right)\left(\left|x\right|+9\right)\) (*)
Th1: x >/ 0
(*) \(\Leftrightarrow x^2+5x+9=\left(x+5\right)\left(x+9\right)\)
\(\Leftrightarrow x^2+5x+9=x^2+14x+45\)
\(\Leftrightarrow9x=36\Leftrightarrow x=4\left(N\right)\)
Th2: x \< 0
(*) \(\Leftrightarrow x^2+5x+9=\left(x+5\right)\left(9-x\right)\)
\(\Leftrightarrow2x^2+x-36=0\Leftrightarrow\left[{}\begin{matrix}x=4\left(L\right)\\x=-\dfrac{9}{2}\left(N\right)\end{matrix}\right.\)
Kl: x=4 , x= - 9/2
c) Đk: \(x\ge-\dfrac{1}{3}\)
\(\sqrt{3x+1}=3x+1\Leftrightarrow\sqrt{3x+1}\left(\sqrt{3x+1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3x+1}=0\\\sqrt{3x+1}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\left(N\right)\\x=0\left(N\right)\end{matrix}\right.\)
Kl: x= -1/3 , x=0
(1)Phương trình đã cho tương đương với:
√3x2−7x+3−√3x2−5x−1=√x2−2−√x2−3x+43x2−7x+3−3x2−5x−1=x2−2−x2−3x+4
⇔−2x+4√3x2−7x+3+√3x2−5x−1=3x−6√x2−2+√x2−3x+4⇔−2x+43x2−7x+3+3x2−5x−1=3x−6x2−2+x2−3x+4
Đến đây thì bạn có thể suy ra nghiệm của phương trình sau cùng là x=2x=2. Kiểm tra lại điều kiện ban đầu để kết luận nghiệm của phương trình đã cho.
(2)đk:23≤x≤723≤x≤7
Phương trình đã cho tương đương với:
3x−18√3x−2+4+x−6√7−x−1+(x−6)(3x2+x−2)3x−183x−2+4+x−67−x−1+(x−6)(3x2+x−2)=0
⇔(x−6)(3√3x−2+4+1√7−x−1+3x2+x−2)⇔(x−6)(33x−2+4+17−x−1+3x2+x−2)=0
⇔x=6⇔x=6
vì với 23≤x≤723≤x≤7
thì: (3√3x−2+4+1√7−x−1+3x2+x−2)(33x−2+4+17−x−1+3x2+x−2)>0
5.
ĐKXĐ: ...
\(\Leftrightarrow3x^2-14x-5+\sqrt{3x+1}-4+1-\sqrt{6-x}=0\)
\(\Leftrightarrow\left(3x+1\right)\left(x-5\right)+\frac{3\left(x-5\right)}{\sqrt{3x+1}+4}+\frac{x-5}{1+\sqrt{6-x}}=0\)
\(\Leftrightarrow\left(x-5\right)\left(3x+1+\frac{3}{\sqrt{3x+1}+4}+\frac{1}{1+\sqrt{6-x}}\right)=0\)
\(\Leftrightarrow x=5\)
6.
ĐKXĐ: \(-4\le x\le4\)
\(\Leftrightarrow\frac{\left(\sqrt{x+4}-2\right)\left(\sqrt{x+4}+2\right)\left(\sqrt{4-x}+2\right)}{\sqrt{x+4}+2}=2x\)
\(\Leftrightarrow\frac{x\left(\sqrt{4-x}+2\right)}{\sqrt{x+4}+2}=2x\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\frac{\sqrt{4-x}+2}{\sqrt{x+4}+2}=2\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{4-x}+2=2\sqrt{x+4}+4\)
\(\Leftrightarrow2\sqrt{x+4}-\frac{4}{5}+\frac{14}{5}-\sqrt{4-x}=0\)
\(\Leftrightarrow\frac{2\left(x+4-\frac{4}{25}\right)}{\sqrt{x+4}+\frac{2}{5}}+\frac{\frac{196}{25}-4+x}{\frac{14}{5}+\sqrt{4-x}}=0\)
\(\Leftrightarrow\left(x-\frac{96}{25}\right)\left(\frac{2}{\sqrt{x+4}+\frac{2}{5}}+\frac{1}{\frac{14}{5}+\sqrt{4-x}}\right)=0\)
\(\Rightarrow x=\frac{96}{25}\)
1.
Bạn coi lại đề
2.
ĐKXĐ: \(1\le x\le2\)
Nhận thấy \(\sqrt{x+2}+\sqrt{x-1}>0;\forall x\) , nhân 2 vế của pt với nó:
\(\left(\sqrt{x+2}+\sqrt{x-1}\right)\left(\sqrt{x+2}-\sqrt{x-1}\right)\left(\sqrt{2-x}+1\right)=\sqrt{x+2}+\sqrt{x-1}\)
\(\Leftrightarrow3\left(\sqrt{2-x}+1\right)=\sqrt{x+2}+\sqrt{x-1}\)
\(\Leftrightarrow3\sqrt{2-x}+3=\sqrt{x+2}+\sqrt{x-1}\)
\(\Leftrightarrow3\sqrt{2-x}+2-\sqrt{x+2}+1-\sqrt{x-1}=0\)
\(\Leftrightarrow3\sqrt{2-x}+\frac{2-x}{2+\sqrt{x+2}}+\frac{2-x}{1+\sqrt{x-1}}=0\)
\(\Leftrightarrow\sqrt{2-x}\left(3+\frac{\sqrt{2-x}}{2+\sqrt{x+2}}+\frac{\sqrt{2-x}}{1+\sqrt{x-1}}\right)=0\)
\(\Leftrightarrow\sqrt{2-x}=0\Rightarrow x=2\)
\(A\sqrt{3}=\sqrt{3x+6\sqrt{3x-9}}+\sqrt{3x-6\sqrt{3x-9}}\)
\(=\sqrt{3x-9+6\sqrt{3x-9}+9}+\sqrt{3x-9-6\sqrt{3x-9}+9}\)
\(=\sqrt{\left(\sqrt{3x-9}+3\right)^2}+\sqrt{\left(\sqrt{3x-9}-3\right)^2}\)
\(=\left|\sqrt{3x-9}+3\right|+\left|\sqrt{3x-9}-3\right|\)
Do \(x\ge6\Rightarrow\sqrt{3x-9}-3\ge0\)
\(\Rightarrow A\sqrt{3}=\sqrt{3x-9}+3+\sqrt{3x-9}-3=2\sqrt{3x-9}\ge6\)
\(\Rightarrow A\ge\frac{6}{\sqrt{3}}=2\sqrt{3}\)
Dấu "=" xảy ra khi \(x=6\)
\(2M=\left(\sqrt{x^2-3x+25}-\sqrt{x^2-3x+9}\right)\)\(\left(\sqrt{x^2-3x+25}+\sqrt{x^2-3x+9}\right)\)
\(2M=x^2-3x+25-x^2+3x-9=16\)
M = 8