Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
CÂU 1:
\(\dfrac{6x^2y^2}{8xy^5}=\dfrac{3x}{4y^3}\)
CÂU 2:
\(\dfrac{12x^3y^2}{18xy^5}=\dfrac{2x^2}{3y^3}\)
CÂU 3:
\(\dfrac{15x\left(x+5\right)^3}{20x^2\left(x+5\right)}=\dfrac{3\left(x+5\right)^2}{4x}\)
CÂU 4:
\(\dfrac{3xy+x}{9y+3}=\dfrac{x\left(3y+1\right)}{3\left(3y+1\right)}=\dfrac{x}{3}\)
CÂU 5:
\(\dfrac{3xy+3x}{9y+9}=\dfrac{3x\left(y+1\right)}{9\left(y+1\right)}=\dfrac{x}{3}\)
CÂU 6:
\(\dfrac{x^2-xy}{5y^2-5xy}=\dfrac{x\left(x-y\right)}{5y\left(y-x\right)}=\dfrac{-x\left(y-x\right)}{5y\left(y-x\right)}=\dfrac{-x}{5y}\)
CÂU 7:
\(\dfrac{2x^2+2x}{x+1}=\dfrac{2x\left(x+1\right)}{x+1}=2x\)
CÂU 8:
\(\dfrac{7x^2+14x+7}{3x^2+3x}=\dfrac{7\left(x^2+2x+1\right)}{3x\left(x+1\right)}\\ =\dfrac{7\left(x+1\right)^2}{3x\left(x+1\right)}=\dfrac{7\left(x+1\right)}{3x}\)
CÂU 9:
\(\dfrac{10xy^2\left(x+y\right)}{15xy\left(x+y\right)^3}=\dfrac{2y}{3\left(x+y\right)^2}\)
a: BC=căn 6^2+8^2=10cm
bD là phân giác
=>AD/AB=CD/BC
=>AD/3=CD/5=(AD+CD)/(3+5)=8/8=1
=>AD=3cm; CD=5cm
b: Xét ΔBHA vuông tại H và ΔBAC vuông tại A có
góc B chung
=>ΔBHA đồng dạng với ΔBAC
=>BH/BA=BA/BC
=>BH*BC=BA^2
c: Xét ΔBHA có BI là phân giác
nên IH/IA=BH/BA
=>IH/IA=BA/BC=AD/DC
\(A=\left(a^4-2a^3+a^2\right)+\left(a^2-2a+1\right)+1\)
\(A=\left(a^2-a\right)^2+\left(a-1\right)^2+1\ge1\)
\(A_{min}=1\) khi \(\left\{{}\begin{matrix}a^2-a=0\\a-1=0\end{matrix}\right.\) \(\Rightarrow a=1\)
Bài 1:
a) \(=10x\left(x+y\right)+5\left(x+y\right)=5\left(x+y\right)\left(2x+1\right)\)
b) \(=a\left(5y+x\right)-3b\left(5y+x\right)=\left(5y+z\right)\left(a-3b\right)\)
c) \(=x^2\left(x+1\right)-\left(x+1\right)=\left(x+1\right)\left(x^2-1\right)\)
\(=\left(x+1\right)\left(x+1\right)\left(x-1\right)=\left(x+1\right)^2\left(x-1\right)\)
Bài 2:
a) \(\Rightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
b) \(\Rightarrow x^2\left(x+1\right)+\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x^2+1\right)=0\)
\(\Rightarrow x=-1\)(do \(x^2+1>0\))
c) K hiểu đề lắm
d) \(\Rightarrow2x\left(3x-5\right)+2\left(3x-5\right)=0\)
\(\Rightarrow2\left(3x-5\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-1\end{matrix}\right.\)
\(1,\left(x+2\right)\left(3x-4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+2=0\\3x-4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{4}{3}\end{matrix}\right.\)
\(2,\dfrac{x-51}{9}+\dfrac{x-52}{8}=\dfrac{x-53}{7}+\dfrac{x-54}{6}\\ \Leftrightarrow\left(\dfrac{x-51}{9}-1\right)+\left(\dfrac{x-52}{8}-1\right)=\left(\dfrac{x-53}{7}-1\right)+\left(\dfrac{x-54}{6}-1\right)\\ \Leftrightarrow\dfrac{x-60}{9}+\dfrac{x-60}{8}-\dfrac{x-60}{7}-\dfrac{x-60}{6}=0\\ \Leftrightarrow\left(x-60\right)\left(\dfrac{1}{9}+\dfrac{1}{8}-\dfrac{1}{7}-\dfrac{1}{6}\right)=0\)
Vì \(\dfrac{1}{9}+\dfrac{1}{8}-\dfrac{1}{7}-\dfrac{1}{6}\ne0\Rightarrow x-60=0\Rightarrow x=60\)
3,ĐKXĐ:\(x\ne\pm2\)
\(\dfrac{x-2}{x+2}+\dfrac{3}{x-2}=\dfrac{x^2-11}{x^2-4}\\ \Leftrightarrow\left(x-2\right)^2+3\left(x+2\right)=x^2-11\\ \Leftrightarrow x^2-4x+4+3x+6-x^2+11=0\)
\(\Leftrightarrow-x+21=0\\ \Leftrightarrow-x=-21\\ \Leftrightarrow x=21\left(tm\right)\)
\(4,\dfrac{x^2+1}{2}=\dfrac{2x^2+x}{3}\\ \Leftrightarrow3\left(x^2+1\right)=2\left(2x^2+x\right)\\ \Leftrightarrow3x^2+3=4x^2+2x\\ \Leftrightarrow4x^2+2x-3x^2-3=0\\ \Leftrightarrow x^2+2x-3=0\\ \Leftrightarrow\left(x^2+3x\right)-\left(x+3\right)=0\\ \Leftrightarrow x\left(x+3\right)-\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)
1: =x^2+2x+1+1
=(x+1)^2+1>=1>0 với mọi x
=>A luôn dương
2:C=x^2-x+1/4+3/4
=(x-1/2)^2+3/4>=3/4>0 với mọi x
=>C luôn dương
3: E=x^2+3x+9/4+3/4
=(x+3/2)^2+3/4>=3/4>0 với mọi x
=>E luôn dương
7: G=3(x^2-5/3x+1)
=3(x^2-2*x*5/6+25/36+11/36)
=3(x-5/6)^2+11/12>=11/12>0 với mọi x
=>ĐPCM
9: =4(x^2+3/4x+1/2)
=4(x^2+2*x*3/8+9/64+23/64)
=4(x+3/8)^2+23/16>=23/16>0 với mọi x
Chứng minh rằng gì thế em, chụp đủ đề chứ