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a: Ta có: \(\sqrt{1-x^2}=x-1\)
\(\Leftrightarrow1-x^2=x-1\)
\(\Leftrightarrow1-x^2-x+1=0\)
\(\Leftrightarrow x^2+x-2=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)
b: Ta có: \(\sqrt{x^2+4x+4}=x-2\)
\(\Leftrightarrow\left|x+2\right|=x-2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=x-2\left(x\ge-2\right)\\x+2=2-x\left(x< -2\right)\end{matrix}\right.\Leftrightarrow2x=0\)
hay x=0(loại)
Do \(x^6-x^3+x^2-x+1=\left(x^3-\dfrac{1}{2}\right)^2+\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{2}>0\) ; \(\forall x\) nên BPT tương đương:
\(\sqrt{13}-\sqrt{2x^2-2x+5}-\sqrt{2x^2-4x+4}\ge0\)
\(\Leftrightarrow\sqrt{4x^2-4x+10}+\sqrt{4x^2-8x+8}\le\sqrt{26}\) (1)
Ta có:
\(VT=\sqrt{\left(2x-1\right)^2+3^2}+\sqrt{\left(2-2x\right)^2+2^2}\ge\sqrt{\left(2x-1+2-2x\right)^2+\left(3+2\right)^2}=\sqrt{26}\) (2)
\(\Rightarrow\left(1\right);\left(2\right)\Rightarrow\sqrt{4x^2-4x+10}+\sqrt{4x^2-8x+8}=\sqrt{26}\)
Dấu "=" xảy ra khi và chỉ khi \(2\left(2x-1\right)=3\left(2-2x\right)\Leftrightarrow x=\dfrac{4}{5}\)
Vậy BPT có nghiệm duy nhất \(x=\dfrac{4}{5}\)
Câu 4:
Giả sử điều cần chứng minh là đúng
\(\Rightarrow x=y\), thay vào điều kiện ở đề bài, ta được:
\(\sqrt{x+2014}+\sqrt{2015-x}-\sqrt{2014-x}=\sqrt{x+2014}+\sqrt{2015-x}-\sqrt{2014-x}\) (luôn đúng)
Vậy điều cần chứng minh là đúng
2) \(\sqrt{x^2-5x+4}+2\sqrt{x+5}=2\sqrt{x-4}+\sqrt{x^2+4x-5}\)
⇔ \(\sqrt{\left(x-4\right)\left(x-1\right)}-2\sqrt{x-4}+2\sqrt{x+5}-\sqrt{\left(x+5\right)\left(x-1\right)}=0\)
⇔ \(\sqrt{x-4}.\left(\sqrt{x-1}-2\right)-\sqrt{x+5}\left(\sqrt{x-1}-2\right)=0\)
⇔ \(\left(\sqrt{x-4}-\sqrt{x+5}\right)\left(\sqrt{x-1}-2\right)=0\)
⇔ \(\left[{}\begin{matrix}\sqrt{x-4}-\sqrt{x+5}=0\\\sqrt{x-1}-2=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}\sqrt{x-4}=\sqrt{x+5}\\\sqrt{x-1}=2\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x\in\varnothing\\x=5\end{matrix}\right.\)
⇔ x = 5
Vậy S = {5}
1) \(-2x^2+x+1-2\sqrt[]{x^2+x+1}=0\)
\(\Leftrightarrow2\sqrt[]{x^2+x+1}=-2x^2+x+1\left(1\right)\)
Ta có :
\(2\sqrt[]{x^2+x+1}=2\sqrt[]{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\ge\sqrt[]{3}\)
Dấu "=" xảy ra khi và chỉ khi \(x+\dfrac{1}{2}=0\Leftrightarrow x=-\dfrac{1}{2}\)
\(\left(1\right)\Leftrightarrow-2x^2+x+1=\sqrt[]{3}\)
\(\Leftrightarrow2x^2-x+\sqrt[]{3}-1=0\)
\(\Delta=1-8\left(\sqrt[]{3}-1\right)=9-8\sqrt[]{3}\)
\(pt\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt[]{9-8\sqrt[]{3}}}{4}\left(loại\right)\\x=\dfrac{1-\sqrt[]{9-8\sqrt[]{3}}}{4}\left(loại\right)\end{matrix}\right.\) \(\left(vì.x=-\dfrac{1}{2}\right)\)
Vậy phương trình cho vô nghiệm
Lời giải:
a.
PT $\Leftrightarrow |2x+1|=|x-1|$
$\Leftrightarrow 2x+1=x-1$ hoặc $2x+1=-(x-1)$
$\Leftrightarrow x+2=0$ hoặc $3x=0$
$\Leftrightarrow x=-2$ hoặc $x=0$ (tm)
b.
PT $\Leftrightarrow 9x^2-6x+1=x^2-4x+4$
$\Leftrightarrow 8x^2-2x-3=0$
$\Leftrightarrow (4x-3)(2x+1)=0$
$\Leftrightarrow 4x-3=0$ hoặc $2x+1=0$
$\Leftrightarrow x=\frac{3}{4}$ hoặc $x=\frac{-1}{2}$ (tm)
a: =>|2x+1|=|x-1|
=>2x+1=x-1 hoặc 2x+1=-x+1
=>x=-2 hoặc x=0
b: =>|3x-1|=|x-2|
=>3x-1=x-2 hoặc 3x-1=-x+2
=>2x=-1 hoặc 4x=3
=>x=-1/2 hoặc x=3/4
\(ĐK:x\ge\dfrac{1}{2}\\ PT\Leftrightarrow2x-2\sqrt{2x^2+5x-3}=1+x\sqrt{2x-1}-2x\sqrt{x+3}\\ \Leftrightarrow\left(2x-2\right)-\left(2\sqrt{2x^2+5x-3}-4\right)=\left(x\sqrt{2x-1}-x\right)-\left(2x\sqrt{x+3}-4x\right)-3x+3\\ \Leftrightarrow2\left(x-1\right)-\dfrac{2\left(2x^2+5x-7\right)}{\sqrt{2x^2+5x-3}+4}=\dfrac{x\left(2x-2\right)}{\sqrt{2x-1}+1}-\dfrac{2x\left(x-1\right)}{\sqrt{x+3}+4x}-3\left(x-1\right)\\ \Leftrightarrow2\left(x-1\right)-\dfrac{2\left(x-1\right)\left(2x+7\right)}{\sqrt{2x^2+5x-3}+4}-\dfrac{2x\left(x-1\right)}{\sqrt{2x-1}+1}+\dfrac{2x\left(x-1\right)}{\sqrt{x+3}+4x}+3\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left[2-\dfrac{2\left(2x+7\right)}{\sqrt{2x^2+5x-3}+4}-\dfrac{2x}{\sqrt{2x-1}+2}+\dfrac{2x}{\sqrt{x+3}+4x}+3\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\2-\dfrac{2\left(2x+7\right)}{\sqrt{2x^2+5x-3}+4}-\dfrac{2x}{\sqrt{2x-1}+2}+\dfrac{2x}{\sqrt{x+3}+4x}+3=0\left(1\right)\end{matrix}\right.\)
Với \(x\ge\dfrac{1}{2}\Leftrightarrow-\dfrac{2\left(2x+7\right)}{\sqrt{2x^2+5x-3}+4}>-\dfrac{2\cdot8}{4}=-4\)
\(-\dfrac{2x}{\sqrt{2x-1}+2}>-\dfrac{1}{2};\dfrac{2x}{\sqrt{x+3}+4x}>0\)
Do đó \(\left(1\right)>2-4-\dfrac{1}{2}+3=\dfrac{1}{2}>0\) nên (1) vô nghiệm
Vậy PT có nghiệm duy nhất \(x=1\)
Tham khảo:
1) Giải phương trình : \(11\sqrt{5-x}+8\sqrt{2x-1}=24+3\sqrt{\left(5-x\right)\left(2x-1\right)}\) - Hoc24
1.
\(\Leftrightarrow\left(2x+1\right)\sqrt{2x^2+4x+5}-\left(2x+1\right)\left(x+3\right)+x^2-2x-4=0\)
\(\Leftrightarrow\left(2x+1\right)\left(\sqrt{2x^2+4x+5}-\left(x+3\right)\right)+x^2-2x-4=0\)
\(\Leftrightarrow\dfrac{\left(2x+1\right)\left(x^2-2x-4\right)}{\sqrt{2x^2+4x+5}+x+3}+x^2-2x-4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\\dfrac{2x+1}{\sqrt{2x^2+4x+5}+x+3}+1=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2x+1+\sqrt{2x^2+4x+5}+x+3=0\)
\(\Leftrightarrow\sqrt{2x^2+4x+5}=-3x-4\) \(\left(x\le-\dfrac{4}{3}\right)\)
\(\Leftrightarrow2x^2+4x+5=9x^2+24x+16\)
\(\Leftrightarrow7x^2+20x+11=0\)
2.
ĐKXĐ: ...
\(\Leftrightarrow2x\sqrt{2x+7}+7\sqrt{2x+7}=x^2+2x+7+7x\)
\(\Leftrightarrow\left(x^2-2x\sqrt{2x+7}+2x+7\right)+7\left(x-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)^2+7\left(x-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)\left(x+7-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2x+7}\\x+7=\sqrt{2x+7}\end{matrix}\right.\)
\(\Leftrightarrow...\)
