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3 tháng 11 2021
a: Xét tứ giác AECF có
AE//CF
AE=CF
Do đó: AECF là hình bình hành
GH
12 tháng 7 2023
1
Với \(\left\{{}\begin{matrix}x\ne2\\x\ne-1\\x\ne\sqrt{\dfrac{1}{2}}\end{matrix}\right.\)
\(M=\left(\dfrac{x-1}{2-x}-\dfrac{x^2}{x^2-x-2}\right)\left(\dfrac{x^2+2x+1}{4x^4-4x^2+1}\right)\\ =\left(\dfrac{\left(x-1\right)\left(x+1\right)}{\left(2-x\right)\left(x+1\right)}+\dfrac{x^2}{\left(x+1\right)\left(2-x\right)}\right)\left(\dfrac{\left(x+1\right)^2}{\left(2x^2-1\right)^2}\right)\\ =\dfrac{x^2-1+x^2}{\left(x+1\right)\left(2-x\right)}\left(\dfrac{\left(x+1\right)^2}{\left(2x^2-1\right)^2}\right)\\ =\dfrac{\left(2x^2-1\right)\left(x+1\right)^2}{\left(x+1\right)\left(2-x\right)\left(2x^2-1\right)^2}\\ =\dfrac{x+1}{\left(2-x\right)\left(2x^2-1\right)}\)
2
Để M = 0 thì \(\dfrac{x+1}{\left(2-x\right)\left(2x^2-1\right)}=0\Rightarrow x+1=0\Rightarrow x=-1\) (loại)
Vậy không có giá trị x thỏa mãn M = 0
H9
HT.Phong (9A5)
CTVHS
12 tháng 7 2023
1) \(M=\left(\dfrac{x-1}{2-x}-\dfrac{x^2}{x^2-x-2}\right)\cdot\dfrac{x^2+2x+1}{4x^4-4x^2+1}\) (ĐK: \(\left\{{}\begin{matrix}x\ne2\\x\ne-1\\x\ne\sqrt{\dfrac{1}{2}}\end{matrix}\right.\))
\(M=\left(\dfrac{-\left(x-1\right)}{x-2}-\dfrac{x^2}{\left(x-2\right)\left(x+1\right)}\right)\cdot\dfrac{\left(x+1\right)^2}{\left(2x^2-1\right)^2}\)
\(M=\left(\dfrac{-\left(x-1\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}-\dfrac{x^2}{\left(x-2\right)\left(x+1\right)}\right)\cdot\dfrac{\left(x+1\right)^2}{\left(2x^2-1\right)^2}\)
\(M=\left(\dfrac{-\left(x^2-1\right)-x^2}{\left(x-2\right)\left(x+1\right)}\right)\cdot\dfrac{\left(x+1\right)^2}{\left(2x^2-1\right)^2}\)
\(M=\left(\dfrac{-x^2+1-x^2}{\left(x-2\right)\left(x+1\right)}\right)\cdot\dfrac{\left(x+1\right)^2}{\left(2x^2-1\right)^2}\)
\(M=\dfrac{-2x^2+1}{\left(x-2\right)\left(x+1\right)}\cdot\dfrac{\left(x+1\right)^2}{\left(2x^2-1\right)^2}\)
\(M=\dfrac{-\left(2x^2-1\right)\left(x+1\right)^2}{\left(x-2\right)\left(x+1\right)\left(2x^2-1\right)^2}\)
\(M=\dfrac{-\left(x+1\right)}{\left(x-2\right)\left(2x^2-1\right)}\)
2) Ta có: \(M=0\)
\(\Rightarrow\dfrac{-\left(x+1\right)}{\left(x-2\right)\left(2x^2-1\right)}=0\)
\(\Leftrightarrow-\left(x+1\right)=0\)
\(\Leftrightarrow-x=1\)
\(\Leftrightarrow x=-1\left(ktm\right)\)
TM
25 tháng 8 2023
Bài 1. (a) Điều kiện: \(x\ne\pm1\).
Ta có: \(A=\left(\dfrac{x-2}{x-1}-\dfrac{x+3}{x+1}+\dfrac{3}{x-1}\right):\left(1-\dfrac{x+3}{x+1}\right)\)
\(=\left(\dfrac{x-2+3}{x-1}-\dfrac{x+3}{x+1}\right):\dfrac{x+1-\left(x+3\right)}{x+1}\)
\(=\left(\dfrac{x+1}{x-1}-\dfrac{x+3}{x+1}\right):\dfrac{x+1-x-3}{x+1}\)
\(=\dfrac{\left(x+1\right)^2-\left(x+3\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}:\dfrac{-2}{x+1}\)
\(=\dfrac{x^2+2x+1-x^2-2x+3}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{-2}\)
\(=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{-2}=\dfrac{2}{1-x}\)
Vậy: \(A=\dfrac{2}{1-x}\)
(b) \(A=3\Leftrightarrow\dfrac{2}{1-x}=3\)
\(\Rightarrow1-x=\dfrac{2}{3}\Leftrightarrow x=\dfrac{1}{3}\left(TM\right)\)
Vậy: \(x=\dfrac{1}{3}\)
Bài 2. (a) Phương trình tương đương với:
\(\dfrac{3\left(3x-2\right)}{12}+\dfrac{6\left(x+3\right)}{12}=\dfrac{4\left(x-1\right)}{12}+\dfrac{x+1}{12}\)
\(\Rightarrow3\left(3x-2\right)+6\left(x+3\right)=4\left(x-1\right)+x+1\)
\(\Leftrightarrow9x-6+6x+18=4x-4+x+1\)
\(\Leftrightarrow10x=-15\Leftrightarrow x=-\dfrac{3}{2}\)
Vậy: Phương trình có tập nghiệm \(S=\left\{-\dfrac{3}{2}\right\}\).
(b) Điều kiện: \(x\ne\pm1\). Phương trình tương đương với:
\(\dfrac{2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{2\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=\dfrac{2x^2+2}{\left(x+1\right)\left(x-1\right)}\)
\(\Rightarrow2\left(x+1\right)+2\left(x-1\right)=2x^2+2\)
\(\Leftrightarrow2x+2+2x-2=2x^2+2\)
\(\Leftrightarrow2x^2-4x+2=0\Leftrightarrow2\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow2\left(x-1\right)^2=0\Rightarrow x-1=0\Leftrightarrow x=1\left(KTM\right)\)
Vậy: Phương trình có tập nghiệm \(S=\varnothing\)
19 tháng 8 2023
2:
1: =7x(x-y)-5(x-y)
=(x-y)(7x-5)
2: =(x^2-y^2)-(4x-4y)
=(x-y)(x+y)-4(x-y)
=(x-y)(x+y-4)
3: =(x^2+2xy+y^2)-(2x+2y)+1
=(x+y)^2-2(x+y)+1
=(x+y-1)^2
19 tháng 8 2023
3:
1: =>15x-9x+6=45-10x+25
=>6x+6=-10x+70
=>16x=64
=>x=4
2: =>x^2+4x-16-16=0
=>x^2+4x-32=0
=>(x+8)(x-4)=0
=>x=4 hoặc x=-8
3: ĐKXĐ: x<>4; x<>-4
\(PT\Leftrightarrow\dfrac{x+4+\left(x+2\right)\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{5x-4}{\left(x-4\right)\left(x+4\right)}\)
=>x+4+x^2-2x-8=5x-4
=>x^2-x-4=5x-4
=>x^2-6x=0
=>x(x-6)=0
=>x=0 hoặc x=6
4: \(\Leftrightarrow5\left(4x+1\right)-x+2>=3\left(2x-3\right)\)
=>20x+5-x+2>=6x-9
=>19x+7>=6x-9
=>13x>=-16
=>x>=-16/13
mn giup mik bai 3 nay dc ko aj
TM
18 tháng 7 2023
\(D=\dfrac{x^2}{x^2-1}+\dfrac{1}{x^2-x^4}=\dfrac{x^4}{x^2\left(x^2-1\right)}-\dfrac{1}{x^2\left(x^2-1\right)}=\dfrac{x^4-1}{x^2\left(x^2-1\right)}=\dfrac{\left(x^2-1\right)\left(x^2+1\right)}{x^2\left(x^2-1\right)}=\dfrac{x^2+1}{x^2}=1+\dfrac{1}{x^2}\)
do \(x\ne0,\pm1\Rightarrow\dfrac{1}{x^2}>0\Rightarrow1+\dfrac{1}{x^2}>1\Rightarrow D>1\left(đpcm\right)\)
O
18 tháng 7 2023
\(D=\dfrac{x^2}{x^2-1}+\dfrac{1}{x^2-x^4}\\ =\dfrac{x^4\left(1-x\right)}{\left(x-1\right)\left(x+1\right)\left(1-x\right)x^2}+\dfrac{x-1}{x^2\left(1-x\right)\left(1+x\right)\left(x-1\right)}\\ =\dfrac{x^4-x^5+x-1}{x^2\left(1-x\right)\left(1+x\right)\left(x-1\right)}\\ =\dfrac{-\left(x-1\right)^2\left(x^2+1\right)\left(x+1\right)}{-x^2\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{x^2+1}{x^2}>1\left(đpcm\right)\)
(x2 + 1 luôn lớn hơn x2)
HP
4 tháng 2 2022
\(\dfrac{10x+3}{12}=1+\dfrac{6+8x}{9}\)
\(\Leftrightarrow\dfrac{30x+9}{36}=\dfrac{36}{36}+\dfrac{24+32x}{36}\)
\(\Leftrightarrow30x+9=36+24+32x\)
\(\Leftrightarrow-2x=51\)
\(\Leftrightarrow x=-\dfrac{51}{2}\)
28 tháng 7 2023
1" =>-4y-y=-5-13
=>-5y=-18
=>y=18/5
2: =>2x+x=2/3+3/5
=>3x=19/15
=>x=19/45
3: =>17-14x-14=13-5x+15
=>-14x+3=-5x+28
=>-9x=25
=>x=-25/9
4: =>3(3x-7)+2(x+1)=-96
=>9x-21+2x+2=-96
=>11x=-96+19=-77
=>x=-7
5: =>15x-5x-5=6x+3
=>10x-5=6x+3
=>4x=8
=>x=2
1) \(\left(x-2\right)\left(4x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\4x-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\4x=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{3}{2}\end{matrix}\right.\)
2) \(\left(2x^2+5\right)\left(5-10x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2+5=0\\5-10x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2=-5\\10x=5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=-\dfrac{5}{2}\left(\text{vô lí}\right)\\x=\dfrac{1}{2}\end{matrix}\right.\)
3) \(\left(x-3\right)\left(2x+6\right)=\left(4+3x\right)\left(3-x\right)\)
\(\Leftrightarrow\left(x-3\right)\left(2x+6\right)-\left(4+3x\right)\left(3-x\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x+6\right)+\left(4+3x\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left[\left(2x+6\right)+\left(4+3x\right)\right]=0\)
\(\Leftrightarrow\left(x-3\right)\left(5x+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\5x+10=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\5x=-10\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
4) \(\left(4x-3\right)\left(x-5\right)=x^2-16\)
\(\Leftrightarrow\left(4x^2-20x-3x+15\right)-\left(x^2-16\right)=0\)
\(\Leftrightarrow4x^2-23x+15-x^2+16=0\)
\(\Leftrightarrow3x^2-23x+31=0\)
\(\Delta=\left(-23\right)^2-4\cdot3\cdot31=157>0\)
\(\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{-23+\sqrt{157}}{6}\\x_2=\dfrac{-23-\sqrt{157}}{6}\end{matrix}\right.\)
5) \(\left(3x+1\right)^2=x^2-8x+16\)
\(\Leftrightarrow\left(3x+1\right)^2=\left(x-4\right)^2\)
\(\Leftrightarrow\left(3x+1\right)^2-\left(x-4\right)^2=0\)
\(\Leftrightarrow\left[\left(3x+1\right)-\left(x-4\right)\right]\left[\left(3x+1\right)+\left(x-4\right)\right]=0\)
\(\Leftrightarrow\left(2x+5\right)\left(4x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+5=0\\4x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=-5\\4x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\)
1: =>x-2=0 hoặc 4x-6=0
=>x=2 hoặc x=3/2
2: =>5-10x=0
=>10x=5
=>x=1/2
3: =>(x-3)(2x+6)=(x-3)(-3x-4)
=>(x-3)(2x+6+3x+4)=0
=>(x-3)(5x+10)=0
=>x=3 hoặc x=-2
4: =>4x^2-20x-3x+15-x^2+16=0
=>3x^2-23x+31=0
=>\(x=\dfrac{23\pm\sqrt{157}}{6}\)
5: =>(3x+1)^2-(x-4)^2=0
=>(3x+1+x-4)(3x+1-x+4)=0
=>(4x-3)(2x+5)=0
=>x=3/4 hoặc x=-5/2