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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 \(\dfrac{4}{13}< x< \dfrac{11}{10}\)
5,\(hpt\Leftrightarrow\left\{{}\begin{matrix}x\left(x+y\right)\left(x+2\right)=0\\2\sqrt{x^2-2y-1}+\sqrt[3]{y^3-14}=x-2\end{matrix}\right.\)
Thay từng TH rồi làm nha bạn
3,\(hpt\Leftrightarrow\left\{{}\begin{matrix}x-y=\frac{1}{x}-\frac{1}{y}=\frac{y-x}{xy}\\2y=x^3+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(1+\frac{1}{xy}\right)=0\\2y=x^3+1\end{matrix}\right.\)
thay nhá
Bài 1:ĐKXĐ: \(2x\ge y;4\ge5x;2x-y+9\ge0\)\(\Rightarrow2x\ge y;x\le\frac{4}{5}\Rightarrow y\le\frac{8}{5}\)
PT(1) \(\Leftrightarrow\left(x-y-1\right)\left(2x-y+3\right)=0\)
+) Với y = x - 1 thay vào pt (2):
\(\frac{2}{3+\sqrt{x+1}}+\frac{2}{3+\sqrt{4-5x}}=\frac{9}{x+10}\) (ĐK: \(-1\le x\le\frac{4}{5}\))
Anh quy đồng lên đê, chắc cần vài con trâu đó:))
+) Với y = 2x + 3...
\(\sqrt{2x-1}\ge0\)
\(\Rightarrow BPT\ge0\) khi
\(3-2x-x^2\ge0\)
\(\Leftrightarrow x^2+2x-3\le0\)
\(\Leftrightarrow\left(x+1\right)^2-4\le0\)
\(\Leftrightarrow\left(x+1\right)^2\le4\)
\(\Leftrightarrow x+1\le2\)
\(\Rightarrow x\le1\)
a: ĐKXĐ: \(\left(2x^2-5x+2\right)\left(x^3+1\right)< >0\)
=>(2x-1)(x-2)(x+1)<>0
hay \(x\notin\left\{\dfrac{1}{2};2;-1\right\}\)
b: ĐKXĐ: x+5<>0
=>x<>-5
c: ĐKXĐ: x4-1<>0
hay \(x\notin\left\{1;-1\right\}\)
d: ĐKXĐ: \(x^4+2x^2-3< >0\)
=>\(x\notin\left\{1;-1\right\}\)
`sqrt{x-2}-2>=sqrt{2x-5}-sqrt{x+1}`
`đk:x>=5/2`
`bpt<=>\sqrt{x-2}+\sqrt{x+1}>=\sqrt{2x-5}+2`
`<=>x-2+x+1+2\sqrt{(x-2)(x+1)}>=2x-5+4+4\sqrt{2x-5}`
`<=>2x-1+2\sqrt{(x-2)(x+1)}>=2x-1+4\sqrt{2x-5}`
`<=>2\sqrt{(x-2)(x+1)}>=4\sqrt{2x-5}`
`<=>sqrt{x^2-x-2}>=2sqrt{2x-5}`
`<=>x^2-x-2>=4(2x-5)`
`<=>x^2-x-2>=8x-20`
`<=>x^2-9x+18>=0`
`<=>(x-3)(x-6)>=0`
`<=>` \(\left[ \begin{array}{l}x \ge 6\\x \le 3\end{array} \right.\)
Kết hợp đkxđ:
`=>` \(\left[ \begin{array}{l}x \ge 6\\\dfrac52 \le x \le 3\end{array} \right.\)
5: ĐKXĐ: \(\frac{x+3}{x-7}>0\)
=>x>7 hoặc x<-3
Ta có: \(\left(x-7\right)\cdot\sqrt{\frac{x+3}{x-7}}=x+4\)
=>\(\sqrt{\left(x+3\right)\left(x-7\right)}=x+4\)
=>\(\begin{cases}x+4\ge0\\ \left(x+3\right)\left(x-7\right)=\left(x+4\right)^2\end{cases}\Rightarrow\begin{cases}x\ge-4\\ x^2-4x-21=x^2+8x+16\end{cases}\)
=>\(\begin{cases}x\ge-4\\ -12x=37\end{cases}\Rightarrow x=-\frac{37}{12}\) (nhận)
6: ĐKXĐ: x>=4
Ta có: \(2\sqrt{x-4}+\sqrt{x-1}=\sqrt{2x-3}+\sqrt{4x-16}\)
=>\(2\sqrt{x-4}+\sqrt{x-1}=\sqrt{2x-3}+2\sqrt{x-4}\)
=>\(\sqrt{2x-3}=\sqrt{x-1}\)
=>2x-3=x-1
=>2x-x=-1+3
=>x=2(loại)
7: ĐKXĐ: x>=1
Ta có: \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\frac{x+3}{2}\)
=>\(\sqrt{x-1+2\cdot\sqrt{x-1}+1}+\sqrt{x-1-2\cdot\sqrt{x-1}\cdot1+1}=\frac{x+3}{2}\)
=>\(\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=\frac{x+3}{2}\)
=>\(\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|=\frac{x+3}{2}\) (1)
TH1: \(\sqrt{x-1}-1\ge0\)
=>\(\sqrt{x-1}\ge1\)
=>x-1>=1
=>x>=2
(1) sẽ trở thành: \(\sqrt{x-1}+1+\sqrt{x-1}-1=\frac{x+3}{2}\)
=>\(2\sqrt{x-1}=\frac{x+3}{2}\)
=>\(4\sqrt{x-1}=x+3\)
=>\(16\left(x-1\right)=\left(x+3\right)^2\)
=>\(x^2+6x+9=16x-16\)
=>\(x^2-10x+25=0\)
=>\(\left(x-5\right)^2=0\)
=>x-5=0
=>x=5(nhận)
TH2: \(\sqrt{x-1}-1<0\)
=>\(\sqrt{x-1}<1\)
=>0<=x-1<1
=>1<=x<2
(1) sẽ trở thành: \(\sqrt{x-1}+1+1-\sqrt{x-1}=\frac{x+3}{2}\)
=>\(\frac{x+3}{2}=2\)
=>x+3=4
=>x=1(nhận)
e: \(\begin{cases}x\left(x+5\right)<4x+2\\ \left(2x-1\right)\left(x+3\right)\ge4x\end{cases}\Rightarrow\begin{cases}x^2+5x-4x-2<0\\ 2x^2+6x-x-3-4x\ge0\end{cases}\)
=>\(\begin{cases}x^2+x-2<0\\ 2x^2+x-3\ge0\end{cases}\Rightarrow\begin{cases}\left(x+2\right)\left(x-1\right)<0\\ 2x^2+3x-2x-3\ge0\end{cases}\)
=>\(\begin{cases}-2
=>-2<x<=-1
f: ĐKXĐ: x∉{1;4;2;5}
Ta có: \(\frac{1}{x^2-5x+4}\le\frac{1}{x^2-7x+10}\)
=>\(\frac{1}{x^2-5x+4}-\frac{1}{x^2-7x+10}\le0\)
=>\(\frac{x^2-7x+10-x^2+5x-4}{\left(x-1\right)\left(x-4\right)\left(x-2\right)\left(x-5\right)}\le0\)
=>\(\frac{-2x+6}{\left(x-1\right)\left(x-4\right)\left(x-2\right)\left(x-5\right)}\le0\)
=>\(\frac{x-3}{\left(x-1\right)\left(x-2\right)\left(x-4\right)\left(x-5\right)}\ge0\)
Đặt \(A=\frac{x-3}{\left(x-1\right)\left(x-2\right)\left(x-4\right)\left(x-5\right)}\)
Đặt x-3=0
=>x=3
Đặt x-1=0
=>x=1
Đặt x-2=0
=>x=2
Đặt x-4=0
=>x=4
Đặt x-5=0
=>x=5
Bảng xét dấu:
Theo bãng xét dấu, ta có: A>=0 khi 1<x<2; 3<=x<4; x>5
d: \(x\left(2x-3\right)\le3x\left(x-1\right)-1\)
=>\(2x^2-3x\le3x^2-3x-1\)
=>\(-x^2\le-1\)
=>\(x^2\ge1\)
=>\(\left[\begin{array}{l}x\ge1\\ x\le-1\end{array}\right.\)
b: ĐKXĐ: (x-3)(x+1)>=0
=>x>=3 hoặc x<=-1
Ta có: \(\sqrt{x^2-2x-3}>2x-3\)
=>\(\begin{cases}x^2-2x-3\ge\left(2x-3\right)^2\\ 2x-3\ge0\end{cases}\Rightarrow\begin{cases}x^2-2x-3-4x^2+12x-9\ge0\\ x\ge\frac32\end{cases}\)
=>\(\begin{cases}-3x^2+10x-12\ge0\\ x\ge\frac32\end{cases}\Rightarrow\begin{cases}x^2-\frac{10}{3}x+4\ge0\\ x\ge\frac32\end{cases}\)
=>\(\begin{cases}x^2-2\cdot x\cdot\frac53+\frac{25}{9}+\frac{11}{9}\ge0\\ x\ge\frac32\end{cases}\Rightarrow\begin{cases}\left(x-\frac53\right)^2+\frac{11}{9}\ge0\left(luônđúng\right)\\ x\ge\frac32\end{cases}\)
=>x>=3/2
kết hợp ĐKXĐ, ta được: x>=3