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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
a: Xét tứ giác AECF có
AE//CF
AE=CF
Do đó: AECF là hình bình hành
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
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: \(M=0\)
\(\Rightarrow\dfrac{-\left(x+1\right)}{\left(x-2\right)\left(2x^2-1\right)}=0\)
\(\Leftrightarrow-\left(x+1\right)=0\)
\(\Leftrightarrow-x=1\)
\(\Leftrightarrow x=-1\left(ktm\right)\)
x/y = 2/5 ⇒ x/2 = y/5
⇒ x/5 = 2y/10
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
x/2 = 2y/10 = (x + 2y)/(2 + 10) = 36/12 = 3
x/2 = 3 ⇒ x = 2 . 3 = 6
y/5 = 3 ⇒ y = 5 . 3 = 15
Vậy x = 6; y = 10
a, vì tam giác ABC vuông tại A , áp dụng định lí pytago ta có
\(AB^2+AC^2=BC^2=>BC=\sqrt{AB^2+AC^2}=\sqrt{6^2+8^2}=10cm\)
b,xét tam giác ABH và tam giác CBA ta có
góc B chung
góc AHB= góc BAC=90 độ
=>tam giác ABH đồng dạng tam giác CBA(góc.góc)
=>\(\dfrac{BC}{AB}=\dfrac{AB}{BH}< =>AB^2=BH.BC\)
c,ta có \(AB^2=BH.BC=>BH=\dfrac{AB^2}{BC}=\dfrac{6^2}{10}=\dfrac{18}{5}cm\)
\(=>HC=BC-HB=10-\dfrac{18}{5}=\dfrac{32}{5}\)
Tính giá trị của $x+y-2=0$ là sao nhỉ? $x+y-2=0$ sẵn rồi mà bạn?
a: BD là phân giác của góc ABC
=>\(\widehat{ABD}=\dfrac{1}{2}\cdot\widehat{ABC}\left(1\right)\)
CE là phân giác của góc ACB
=>\(\widehat{ACE}=\dfrac{1}{2}\cdot\widehat{ACB}\left(2\right)\)
ΔABC cân tại A
=>\(\widehat{ABC}=\widehat{ACB}\left(3\right)\)
Từ (1),(2),(3) suy ra \(\widehat{ABD}=\widehat{ACE}\)
Xét ΔABD và ΔACE có
\(\widehat{ABD}=\widehat{ACE}\)
AB=AC
\(\widehat{BAD}\) chung
Do đó: ΔABD=ΔACE
b: ΔABD=ΔACE
=>AD=AE
Xét ΔABC có \(\dfrac{AD}{AC}=\dfrac{AE}{AB}\)
nên DE//BC
Xét tứ giác BEDC có DE//BC
nên BEDC là hình thang
Hình thang BEDC có \(\widehat{EBC}=\widehat{DCB}\)
nên BEDC là hình thang cân
a) Xét \(\Delta MNP\) vuông tại M:
\(NP^2=MN^2+MP^2\left(Pytago\right).\\ \Rightarrow NP^2=5^2+12^2.\\ \Rightarrow NP^2=169.\\ \Rightarrow NP=13\left(cm\right).\)
b) Xét \(\Delta MNP\) vuông tại M:
MD là phân giác \(\widehat{NMP}\left(gt\right).\)
\(\Rightarrow\dfrac{ND}{DP}=\dfrac{MN}{MP}\) (T/c phân giác).
\(\Leftrightarrow\dfrac{ND}{ND+DP}=\dfrac{MN}{MP+MN}.\\ \Leftrightarrow\dfrac{ND}{NP}=\dfrac{MN}{MP+MN}.\\ \Rightarrow\dfrac{ND}{13}=\dfrac{5}{12+5}.\\ \Leftrightarrow\dfrac{ND}{13}=\dfrac{5}{17}.\\ \Rightarrow ND=\dfrac{65}{17}\left(cm\right).\\ \Rightarrow DP=NP-ND=13-\dfrac{65}{17}=\dfrac{156}{17}\left(cm\right).\)