mn giup mik nha plssss
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Những câu hỏi liên quan
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: \(M=0\)
\(\Rightarrow\dfrac{-\left(x+1\right)}{\left(x-2\right)\left(2x^2-1\right)}=0\)
\(\Leftrightarrow-\left(x+1\right)=0\)
\(\Leftrightarrow-x=1\)
\(\Leftrightarrow x=-1\left(ktm\right)\)
TG
3
DH
17 tháng 2 2021
Áp dụng định lí pytago trong \(\Delta MNP\) vuông tại \(M\) có:
\(\Rightarrow MP=\sqrt{NP^2-MN^2}=\sqrt{7,5^2-4,5^2}=6cm\)
\(\Rightarrow D\)
TT
28
DL
7
anime nek bn nhé :)))))
đâu bn gì ơi
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