Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
lời giải
a) \(\left\{{}\begin{matrix}-2x+\dfrac{3}{5}>\dfrac{2x-7}{3}\left(1\right)\\x-\dfrac{1}{2}< \dfrac{5\left(3x-1\right)}{2}\left(2\right)\end{matrix}\right.\)
(1)\(\Leftrightarrow\)
\(\dfrac{3}{5}+\dfrac{7}{3}>\left(\dfrac{2}{3}+2\right)x\)
\(\dfrac{44}{15}>\dfrac{8}{3}x\) \(\Rightarrow x< \dfrac{44.3}{15.8}=\dfrac{11}{5.2}=\dfrac{11}{10}\)
Nghiêm BPT(1) là \(x< \dfrac{11}{10}\)
(2) \(\Leftrightarrow2x-1< 15x-5\Rightarrow13x>4\Rightarrow x>\dfrac{4}{13}\)
Ta có: \(\dfrac{4}{13}< \dfrac{11}{10}\) => Nghiệm hệ (a) là \(\dfrac{4}{13}< x< \dfrac{11}{10}\)
- Thay từng giá trị vào, ta thấy A. \(\dfrac{15}{4}\) thỏa mãn.
Lời giải:
a.
\(\left\{\begin{matrix} x\neq 0\\ 2x-1\geq 0\\ x^2-3x+2=(x-1)(x-2)\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\neq 0\\ x\geq \frac{1}{2}\\ x\neq 1; x\neq 2\end{matrix}\right.\)
$\Leftrightarrow x\geq \frac{1}{2}; x\neq 1; x\neq 2$
b. \(\left\{\begin{matrix}
x^2-1=(x-1)(x+1)\neq 0\\
7-2x\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}
x\neq \pm 1\\
x\leq \frac{7}{2}\end{matrix}\right.\)
c.
\(\left\{\begin{matrix} x\neq 0\\ 4-2x+x^2\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\neq 0\\ (x-1)^2+3\neq 0\end{matrix}\right.\Leftrightarrow x\neq 0\)
d.
\(\left\{\begin{matrix} 25-x^2=(5-x)(5+x)\geq 0\\ x\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} -5\leq x\leq 5\\ x\geq 0\end{matrix}\right.\Leftrightarrow 0\leq x\leq 5\)
a) \(y=\dfrac{1}{x}-\dfrac{\sqrt[]{2x-1}}{x^2-3x+2}\)
Điều kiện \(\) \(2x-1\ge0;x\ne0;x^2-3x+2\ne0\)
\(\Leftrightarrow x\ge\dfrac{1}{2};x\ne0;\left(x-1\right)\left(x-2\right)\ne0\)
\(\Leftrightarrow x\ge\dfrac{1}{2};x\ne0;x\ne1;x\ne2\)
1.
\(-4\le\dfrac{x^2-2x-7}{x^2+1}\le1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-2x-7\le x^2+1\\-4x^2-4\le x^2-2x-7\end{matrix}\right.\) (Do \(x^2+1>0\))
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-4\\\left[{}\begin{matrix}x\ge1\\x\le-\dfrac{3}{5}\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge1\\-4\le x\le-\dfrac{3}{5}\end{matrix}\right.\)
2.
\(\dfrac{1}{13}\le\dfrac{x^2-2x-2}{x^2-5x+7}\le1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-5x+7\le13x^2-26x-26\\x^2-2x-2\le x^2-5x+7\end{matrix}\right.\) (Do \(x^2-5x+7>0\))
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge\dfrac{11}{4}\\x\le-1\end{matrix}\right.\\x\le3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{11}{4}\le x\le3\\x\le-1\end{matrix}\right.\)
1.
ĐK: \(x\ne3;x\ne-2\)
\(\dfrac{5}{x-3}+\dfrac{3}{x+2}\le\dfrac{3+2x}{x^2-x-6}\)
\(\Leftrightarrow\dfrac{5\left(x+2\right)+3\left(x-3\right)}{x^2-x-6}\le\dfrac{3+2x}{x^2-x-6}\)
\(\Leftrightarrow\dfrac{8x+1-3-2x}{x^2-x-6}\le0\)
\(\Leftrightarrow\dfrac{6x-2}{x^2-x-6}\le0\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x-2\ge0\\x^2-x-6< 0\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}6x-2\le0\\x^2-x-6>0\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}6x-2\ge0\\x^2-x-6< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{3}\\-2< x< 3\end{matrix}\right.\Leftrightarrow\dfrac{1}{3}\le x< 3\)
TH2: \(\left\{{}\begin{matrix}6x-2\le0\\x^2-x-6>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{1}{3}\\\left[{}\begin{matrix}x>3\\x< -2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow x< -2\)
Vậy ...
2.
ĐK: \(x\ne\pm2\)
\(\dfrac{1}{x^2-4}+\dfrac{2}{x+2}>-\dfrac{3}{x-2}\)
\(\Leftrightarrow\dfrac{1}{x^2-4}+\dfrac{2\left(x-2\right)+3\left(x+2\right)}{x^2-4}>0\)
\(\Leftrightarrow\dfrac{5x+3}{x^2-4}>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}5x+3>0\\x^2-4>0\end{matrix}\right.\\\left\{{}\begin{matrix}5x+3< 0\\x^2-4< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-\dfrac{3}{5}< x< 2\\x< -2\end{matrix}\right.\)
Vậy ...
a) \(\dfrac{5+x}{4-x}=\dfrac{1}{2}\)
\(\Leftrightarrow2\left(5+x\right)=4-x\)
\(\Leftrightarrow2\left(5+x\right)-\left(4-x\right)=0\)
\(\Leftrightarrow10+2x-4+x=0\)
\(\Leftrightarrow6+3x=0\)
\(\Leftrightarrow3x=-6\)
\(\Leftrightarrow x=-2\)
Vậy x=-2
b) \(\dfrac{25}{14}=\dfrac{x+7}{x-4}\)
\(\Leftrightarrow25\left(x-4\right)=14\left(x+7\right)\)
\(\Leftrightarrow25\left(x-4\right)-14\left(x+7\right)=0\)
\(\Leftrightarrow25x-100-14x-98=0\)
\(\Leftrightarrow11x-198=0\)
\(\Leftrightarrow11x=198\)
\(\Leftrightarrow x=18\)
Vậy x=18
c) \(\dfrac{3x-5}{x+4}=\dfrac{5}{2}\)
\(\Leftrightarrow2\left(3x-5\right)=5\left(x+4\right)\)
\(\Leftrightarrow2\left(3x-5\right)-5\left(x+4\right)=0\)
\(\Leftrightarrow6x-10-5x-20=0\)
\(\Leftrightarrow x-30=0\)
\(\Leftrightarrow x=30\)
Vậy x=30
d) \(\dfrac{3x-1}{2x+1}=\dfrac{3}{7}\)
\(\Leftrightarrow7\left(3x-1\right)=3\left(2x+1\right)\)
\(\Leftrightarrow7\left(3x-1\right)-3\left(2x+1\right)=0\)
\(\Leftrightarrow21x-7-6x-3=0\)
\(\Leftrightarrow15x-10=0\)
\(\Leftrightarrow15x=10\)
\(\Leftrightarrow x=\dfrac{10}{15}=\dfrac{2}{3}\)
Vậy \(x=\dfrac{2}{3}\)