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Tìm x biết:
b/\(\left(2x+3\right)^2-\left(5x-4\right)\left(5x+4\right)=\left(x+5\right)^2-\left(3x-1\right)\left(7x+2\right)-\left(x^2-x+1\right)\)
<=> \(4x^2 +12x+9-25x^2+16-x^2-10x-25+21x^2+6x-7x-2+x^2-x+1=0\)
<=>0x-1=0
<=>0x=1 (vô lí) (dòng này không cần ghi thêm cũng được)
=> Không có giá trị x nào thỏa mãn
c/ \((1-3x)^2-(x-2)(9x+1)=(3x-4)(3x+4)-9(x+3)^2\)
<=>\(1-6x+9x^2-9x^2-x+18x+2-9x^2+16+9x^2+54x+81=0\)
<=> 65x+100=0
<=> x=\(\dfrac{-20}{13}\)
d/\((3x+4)(3x-4)-(2x+5)^2=(x-5)^2+(2x+1)^2-(x^2-2x)+(x-1)^2\)
<=> \(9x^2-16-4x^2-20x-25-x^2+10x-25-4x^2-4x-1+x^2+2x-x^2+2x-1=0\)
<=> -10x-68=0
<=> x=\(\dfrac{-34}{5}\)
\(1,\left|2x-3\right|=x-5\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-5\ge0\\\left[{}\begin{matrix}2x-3=x-5\\2x-3=-x+5\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\text{≥}5\\\left[{}\begin{matrix}x=-2\\x=\frac{8}{3}\end{matrix}\right.\end{matrix}\right.\) (ko thỏa mãn)
=> pt vô nghiệm
\(2,\left|3x+2\right|=x+1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1\text{≥}0\\\left[{}\begin{matrix}3x+2=x+1\\3x+2=-x-1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\text{≥}-1\\\left[{}\begin{matrix}x=-\frac{1}{2}\\x=-\frac{3}{4}\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{2}\\x=-\frac{3}{4}\end{matrix}\right.\)
\(3,\left|2x+1\right|=7-x\)
\(\Leftrightarrow\left\{{}\begin{matrix}7-x\text{≥}0\\\left[{}\begin{matrix}2x+1=7-x\\2x+1=x-7\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\text{≥}7\\\left[{}\begin{matrix}x=2\\x=-8\end{matrix}\right.\end{matrix}\right.\) (loại)
=> pt vô nghiệm
\(4,\left|2x-5\right|=x+1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1\text{≥}0\\\left[{}\begin{matrix}2x-5=x+1\\2x-5=-x-1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\text{≥}-1\\\left[{}\begin{matrix}x=6\\x=\frac{4}{3}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\frac{4}{3}\end{matrix}\right.\)
\(5,\left|6x-2\right|=3x-4\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x-4\text{≥}0\\\left[{}\begin{matrix}6x-2=3x-4\\6x-2=-3x+4\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\text{≥}\frac{4}{3}\\\left[{}\begin{matrix}x=-\frac{2}{3}\\x=\frac{2}{3}\end{matrix}\right.\end{matrix}\right.\) => pt vô nghiệm
\(6,\left|3x-2\right|=x-2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2\text{≥}0\\\left[{}\begin{matrix}3x-2=x-2\\3x-2=-x+2\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\text{≥}2\\\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\end{matrix}\right.\) => pt vô nghiệm
\(7,\left|2x+3\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=1\\2x+3=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)
\(8,\left|2-x\right|=2x-1\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-1\ge0\\\left[{}\begin{matrix}2-x=2x-1\\2-x=-2x+1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{1}{2}\\\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow x=1\)
\(9,\left|2x-1\right|=x-3\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3\ge0\\\left[{}\begin{matrix}2x-1=x-3\\2x-1=-x+3\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge3\\\left[{}\begin{matrix}x=-2\\x=\frac{4}{3}\end{matrix}\right.\end{matrix}\right.\) => pt vô nghiệm
\(10,2\left|x-1\right|=x+2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+2\ge0\\\left[{}\begin{matrix}2x-2=x+2\\2x-2=-x-2\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\)
hoc gioi the hihiihihihhhihihihihiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
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b/ ĐKXĐ: \(-2\le x\le\frac{5}{2}\)
\(\Leftrightarrow\sqrt{x+2}< \sqrt{3-x}+\sqrt{5-2x}\)
\(\Leftrightarrow x+2< 8-3x+2\sqrt{2x^2-11x+6}\)
\(\Leftrightarrow\sqrt{2x^2-11x+6}>2x-3\)
- Với \(x< \frac{3}{2}\Rightarrow\left\{{}\begin{matrix}VT\ge0\\VP< 0\end{matrix}\right.\) BPT luôn đúng
- Với \(x\ge\frac{3}{2}\) hai vế ko âm, bình phương 2 vế:
\(\Leftrightarrow2x^2-11x+6>4x^2-12x+9\)
\(\Leftrightarrow2x^2-x+3< 0\) (vô nghiệm)
Vậy nghiệm của BPT đã cho là:
\(-2\le x< \frac{3}{2}\)
a/ ĐKXĐ: \(\left[{}\begin{matrix}x\le-5\\x\ge-\frac{4}{3}\end{matrix}\right.\)
- Với \(x< 1\Rightarrow\left\{{}\begin{matrix}VT\ge0\\VP< 0\end{matrix}\right.\) BPT luôn đúng
- Với \(x\ge1\) hai vế ko âm, bình phương:
\(\left(x+5\right)\left(3x+4\right)\ge16\left(x-1\right)^2\)
\(\Leftrightarrow3x^2+19x+20\ge16x^2-32x+16\)
\(\Leftrightarrow13x^2-51x-4\le0\Rightarrow-\frac{1}{13}\le x\le4\)
\(\Rightarrow1\le x\le4\)
Kết hợp 2 trường hợp và ĐKXĐ ta được nghiệm của BPT đã cho là:
\(\left[{}\begin{matrix}x\le-5\\-\frac{4}{3}\le x\le4\end{matrix}\right.\)
Đúng rồi đó bạn