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(x-a)(x-b) + (x-b)(x-c) + (x-c)(x-a) + x2
= (x2-ax-bx+ab) + (x2-bx-cx+bc) + (x2-cx-ax+ac) + x2
= 4x2 - 2ax - 2bx + ab + bc + ac
Thay a+b+c = 2x, ta được:
M = 4x2 - 2x(a+b+c) + ab + bc + ac
M = 4x2 - 2x.2x + ab + bc + ac
M = ab + bc + ac
Vậy => đcpcm
Bài 1 rút gọn bc tự làm :
\(B=\dfrac{3y^3-7y^2+5y-1}{2y^3-y^2-4y+3}\)
\(B=\dfrac{3x^3-3y^2-4y^2+4y+y-1}{2y^3-2y^2+y^2-y+3y-3}\)
\(B=\dfrac{3y^2\left(y-1\right)-4y\left(y-1\right)+\left(y-1\right)}{2y^2\left(y-1\right)+y\left(y-1\right)-3\left(y-1\right)}\)
\(B=\dfrac{\left(3y^2-4y+1\right)\left(y-1\right)}{\left(2y^2+y-3\right)\left(y-1\right)}\)
\(B=\dfrac{3y^2-3y-y+1}{2y^2-2y+3y-3}=\dfrac{3y\left(y-1\right)-\left(y-1\right)}{2y\left(y-1\right)+3\left(y-1\right)}\)
\(B=\dfrac{\left(3y-1\right)\left(y-1\right)}{\left(3y+2\right)\left(y-1\right)}=\dfrac{3y-1}{3y+2}\)
Bài 2 )
a ) \(x+\dfrac{1}{x}=3\)
\(\Leftrightarrow x^2+2x\dfrac{1}{x}+\dfrac{1}{x^2}=9\)
\(\Leftrightarrow x^2+\dfrac{1}{x^2}=1\)
b ) \(\left(x+\dfrac{1}{x}\right)^3=27\)
\(\Leftrightarrow x^3+\dfrac{1}{x^3}+\dfrac{3}{x}+3x=27\)
\(\Leftrightarrow x^3+\dfrac{1}{x^3}+3\left(\dfrac{1}{x}+x\right)=27\)
\(\Leftrightarrow x^3+\dfrac{1}{x^3}=18\)
Từ x=\(\dfrac{1}{2}\)a+\(\dfrac{1}{2}\)b+\(\dfrac{1}{2}\)c=\(\dfrac{1}{2}\).(a+b+c)\(\Rightarrow\)2x=(a+b+c)
M=(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)+x\(^2\)
= x\(^2\)-xb-ax+ab+x\(^2\)-xc-bx+bc+x\(^2\)-ax-cx+ac+x\(^2\)
= 4x\(^2\)-2ac-2bx-2cx+ab+bc+ac
= 4x\(^2\)-2x(a+b+c)+ab+bc+ca
Thay 2x=a+b+c,ta được:
M= 4x\(^2\)-2x.2c+ab+bc+ca
M= 4x\(^2\)-4x\(^2\)+ab+bc+ca
M= ab+bc+ca
Câu 3:
\(\Leftrightarrow3x^3-2x^2+6x^2-4x+9x-6>0\)
\(\Leftrightarrow\left(3x-2\right)\left(x^2+2x+3\right)>0\)
=>3x-2>0
=>x>2/3
Câu 1:
a: \(A=x-2+\dfrac{6x-3}{x\left(x+2\right)}+\left(\dfrac{x+1+2x-2}{\left(x^2-1\right)}-\dfrac{3}{x}\right)\cdot\dfrac{x^2-1}{x+2}\)
\(=x-2+\dfrac{6x-3}{x\left(x+2\right)}+\left(\dfrac{3x-1}{x^2-1}-\dfrac{3}{x}\right)\cdot\dfrac{x^2-1}{x+2}\)
\(=x-2+\dfrac{6x-3}{x\left(x+2\right)}+\dfrac{3x^2-x-3x^2+3}{x\left(x^2-1\right)}\cdot\dfrac{x^2-1}{x+2}\)
\(=x-2+\dfrac{6x-3}{x\left(x+2\right)}+\dfrac{-\left(x-3\right)}{x\left(x+2\right)}\)
\(=x-2+\dfrac{6x-3-x^2+3x}{x\left(x+2\right)}\)
\(=x-2+\dfrac{-x^2+9x-3}{x\left(x+2\right)}\)
\(=\dfrac{x\left(x^2-4\right)-x^2+9x-3}{x\left(x+2\right)}\)
\(=\dfrac{x^3-4x-x^2+9x-3}{x\left(x+2\right)}\)
\(=\dfrac{x^3-x^2+5x-3}{x\left(x+2\right)}\)
b: TH1: \(\left\{{}\begin{matrix}x^3-x^2+5x-3>0\\x\left(x+2\right)< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-2< x< 2\\x>0.63\end{matrix}\right.\Leftrightarrow0.63< x< 2\)
TH2: \(\left\{{}\begin{matrix}x^3-x^2+5x-3< 0\\x\left(x+2\right)>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 0.63\\\left[{}\begin{matrix}x>0\\x< -2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0< x< 0.63\\x< -2\end{matrix}\right.\)
\(A=\left(\dfrac{1}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{6x+3}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{2\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\right):\left(x+2\right)\)\(A=\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x^2-x+1\right)\left(x+2\right)}\)
a) \(A=\left\{{}\begin{matrix}x\ne-1;-2\\\dfrac{1}{x^2-x+1}\end{matrix}\right.\)
b)
\(A>1;\dfrac{1}{x^2-x+1}>1\Leftrightarrow x^2-x< 0\Leftrightarrow0< x< 1\)
\(P=\dfrac{1}{x^2-x+1}.\dfrac{x^3-x^2+x}{\left(x+1\right)^2}=\dfrac{x}{\left(x+1\right)^2}\)
x>0 => P >0 đang tìm Giá trị LN => chỉ xét P>0 <=> x>0
\(\dfrac{1}{P}=\dfrac{\left(x+1\right)^2}{x}=x+2+\dfrac{1}{x}\)
áp co si hai số dương x ; 1/x
\(\dfrac{1}{P}\ge2.\sqrt{x.\dfrac{1}{x}}+2=4\Rightarrow P\le\dfrac{1}{4}\)
đẳng thức khi x =1/x => x=1 thỏa mãn đk của x
\(MaxP=\dfrac{1}{4}\)
Câu 1 :
a) Rút gọn P :
\(P=\dfrac{x+1}{3x-x^2}:\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{12x^2}{x^2-9}\right)\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\left[\dfrac{\left(3+x\right)^2}{\left(3-x\right)\left(3+x\right)}-\dfrac{\left(3-x\right)^2}{\left(3-x\right)\left(3+x\right)}-\dfrac{12x^2}{\left(3-x\right)\left(3+x\right)}\right]\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\left(\dfrac{9+6x+x^2-9+6x-x^2-12x^2}{\left(3-x\right)\left(3+x\right)}\right)\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\dfrac{12x-12x^2}{\left(3-x\right)\left(x+3\right)}\)
\(P=\dfrac{x+1}{x\left(3-x\right)}.\dfrac{\left(3-x\right)\left(x+3\right)}{12x\left(1-x\right)}\)
\(P=\dfrac{\left(x+1\right)\left(x+3\right)}{12x^2\left(1-x\right)}\)
a: \(E=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{x+1}{x}+\dfrac{1}{x-1}+\dfrac{2-x^2}{x\left(x-1\right)}\right)\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}\cdot\dfrac{x\left(x-1\right)}{x+1}=\dfrac{x^2}{x-1}\)
b: |x-3|=2
=>x-3=2 hoặc x-3=-2
=>x=5(nhận) hoặc x=1(loại)
Khi x=5 thì \(E=\dfrac{5^2}{5-1}=\dfrac{25}{4}\)
c: Để E=1/2 thì \(\dfrac{x^2}{x-1}=\dfrac{1}{2}\)
\(\Leftrightarrow2x^2-x+1=0\)
hay \(x\in\varnothing\)
f) \(A=\dfrac{x^2}{x-1}=\dfrac{x^2-x+x-1+1}{x-1}=\dfrac{x\left(x-1\right)+x-1+1}{x-1}=x+1+\dfrac{1}{x-1}=x-1+\dfrac{1}{x-1}+2\ge2\sqrt{\left(x-1\right).\dfrac{1}{x-1}}+2=4\)\(A=4\Leftrightarrow x=2\)
-Vậy \(A_{min}=4\)
Bài 2 :
Ta có : \(4p(p-a)\)\(=2\left(a+b+c\right)\left(\dfrac{a+b+c}{2}-a\right)\)
=\(2\left(a+b+c\right)\left(\dfrac{b+c-a}{2}\right)\)
\(=\left(a+b+c\right)\left(b+c-a\right)\)
\(=ab+ac-a^2+b^2+bc-ab+bc+c^2-ac\)
\(=2bc+b^2+c^2-a^2\left(dpcm\right)\)
Vậy :
Bai 2:
Ta có:
\(VP=4p\left(p-a\right)=2p.2p-2a.2p\) (1)
Thay \(a+b+c=2p\) vào (1) ta có:
\(\left(a+b+c\right)^2-2a.\left(a+b+c\right)\)
\(=a^2+b^2+c^2+2ab+2ac+2bc-2a^2-2ab-2ac\)
\(=-a^2+b^2+c^2+2bc=VT\)
Vậy \(2bc+b^2+c^2-a^2=4p\left(p-a\right)\)
Chúc bạn học tốt!!!