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3x2+4y2=7xy
<=> 3x2-3xy+4y2-4xy=0
<=> 3x(x-y)-4y(x-y)=0
<=> (3x-4y)(x-y)=0
<=> 3x-4y=0 hoặc x-y=0
<=> 3x=4y hoặc x=y
<=> y = \(\frac{3}{4}\)x hoặc x=y
+) y = \(\frac{3}{4}\)x, ta có:
F = \(\frac{4.\frac{3}{4}x+2x}{5.\frac{3}{4}x-7x}+\)\(\frac{3x-2.\frac{3}{4}x}{10.\frac{3}{4}x-4x}\)
F = \(\frac{5x}{-\frac{13}{4}x}+\frac{\frac{3}{2}x}{\frac{7}{2}x}\)
F = \(-\frac{20}{13}+\frac{3}{7}=-\frac{101}{91}\)
+) x = y, ta có:
F = \(\frac{4x+2x}{5x-7x}+\frac{3x-2x}{10x-4x}\)
F = \(\frac{6x}{-2x}+\frac{1x}{6x}=-3+\frac{1}{6}=-\frac{17}{6}\)
Từ \(3x^2+4y^2=7xy\Rightarrow3x^2+4y^2-7xy=0\)
\(\Rightarrow3x^2-4xy-3xy+4y^2=0\)
\(\Rightarrow x\left(3x-4y\right)-y\left(3x-4y\right)=0\)
\(\Rightarrow\left(x-y\right)\left(3x-4y\right)=0\)\(\Rightarrow\left[\begin{matrix}x=y\\x=\frac{4y}{3}\end{matrix}\right.\)
*)Xét \(x=y\) ta có \(F=\frac{4y+2y}{5y-7y}+\frac{3y-2y}{10y-4y}=\frac{6y}{-2y}+\frac{y}{6y}=-3+\frac{1}{6}=-\frac{17}{6}\)
*)Xét \(x=\frac{4y}{3}\) ta có \(F=\frac{4y+2\cdot\frac{4y}{3}}{5y-7\cdot\frac{4y}{3}}+\frac{3\cdot\frac{4y}{3}-2y}{10y-4\cdot\frac{4y}{3}}=\frac{4y+\frac{8y}{3}}{5y-\frac{28y}{3}}+\frac{4y-2y}{10y-\frac{16y}{3}}=\frac{-20}{13}+\frac{3}{7}=\frac{-101}{91}\)
a, \(2x^2+3\left(x+1\right)\left(x-1\right)-5x\left(x+1\right)\)
\(=2x^2+3\left(x^2-1\right)-5x^2-5x\)
\(=2x^2+3x^2-3-5x^2-5x\)
\(=\left(2x^2+3x^2-5x^2\right)-3-5x\)
\(=-\left(5x+3\right)\)
b, \(\left(4x+3y\right)\left(2x-5y\right)-\left(2x+6y\right)\left(3x-5y\right)\)
\(=8x^2-20xy+6xy-\left(15y^2-6x^2-10xy-18xy-30y^2\right)\)
\(=8x^2-20xy+6xy-15y^2+6x^2+10xy+18xy+30y^2\)
\(=\left(8x^2+6x^2\right)+\left(-20xy+6xy+10xy+18xy\right)+\left(-15y^2+30y^2\right)\)
\(=14x^2+14xy+15y^2\)
\(=14x.\left(x+y\right)+15y^2\)
Chúc bạn học tốt!!!
1) a) Đặt biểu thức là A
\(A=2x^2+4y^2-4xy-4x-4y+2017\)
\(A=\left(x-2y\right)^2+x^2-4x-4y+2017\)
\(A=\left(x-2y\right)^2+2\left(x-2y\right)+x^2-6x+2017\)
\(A=\left(x-2y-1\right)^2+\left(x+3\right)^2+2008\)
Vậy: MinA=2008 khi x=-3; y=-2
3) a) \(A=\dfrac{1}{x^2+x+1}\)
\(B=x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
\(\Rightarrow B\ge\dfrac{3}{4}\Rightarrow A\ge\dfrac{4}{3}\)
Vậy MinA là \(\dfrac{4}{3}\) khi x=-0,5
1.
a) \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
b) \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
Bài 1:
a, \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow x\left(x+4\right)+\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
Vậy \(x=-4\) hoặc \(x=-1\)
b, \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy \(x=3\) hoặc \(x=-2\)
a) ĐKXĐ : \(x+y\ne0\)
\(x^2-2y^2=xy\)
\(x^2-y^2-y^2-xy=0\)
\(\left(x-y\right)\left(x+y\right)-y\left(y+x\right)=0\)
\(\left(x+y\right)\left(x-2y\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+y=0\left(Loai\right)\\x-2y=0\left(Chon\right)\end{matrix}\right.\)
Với x - 2y = 0 ta có x = 2y
Thay x = 2y vào A ta có :
\(A=\dfrac{2y-y}{2y+y}=\dfrac{y}{3y}=\dfrac{1}{3}\)
\(\dfrac{-5}{6x^5y^3}=\dfrac{-5\cdot2\cdot y^3}{12x^5y^6}=\dfrac{-10y^3}{12x^5y^6}\)
\(\dfrac{3}{4x^2y^6}=\dfrac{3\cdot3x^3}{12x^5y^6}=\dfrac{9x^3}{12x^5y^6}\)
\(\dfrac{7}{3x^4y^5}=\dfrac{7\cdot4\cdot x\cdot y}{12x^5y^6}=\dfrac{28xy}{12x^5y^6}\)
\(A=\left(5x-2y\right)\left(5x+2y\right)\)
\(A=\left(5x\right)^2-\left(2y\right)^2\)
\(A=25x^2-4y^2\)
\(A=25.\left(-2\right)^2-4\left(-10\right)^2\)
\(A=25.4-4.100\)
\(A=100-400\)
\(A=300\)
\(B=\left(2x-5\right)\left(4x^2+10x+25\right)\)
\(B=\left(2x\right)^3-5^3\)
\(B=8x^3-125\)
\(B=8.8-125\)
\(B=64-125\)
\(B=-61\)
\(C=\left(3x+2y\right)\left(9x^2-6xy+4y^2\right)\)
\(C=\left(3x\right)^2+\left(2y\right)^2\)
\(C=9x^2+4y^2\)
\(C=9\left(-1\right)^2+4\left(\dfrac{1}{2}\right)^2\)
\(C=9+4.\dfrac{1}{4}\)
\(C=9+1\)
\(C=10\)
Ta có : 3x2 - 7xy + 4y2 = 0
=> 3x2 - 3xy - 4xy + 4y2 = 0
=> 3x( x - y) - 4y( x - y) = 0
=> ( x - y)( 3x - 4y) = 0
=> x = y ; 3x = 4y
Thay : x = y ; 3x = 4y vào phân thức trên ta có:
\(A=\dfrac{4y+2x}{5y-7x}+\dfrac{3x-2y}{10y-4x}\)
\(A=\dfrac{3x+2x}{5x-7x}+\dfrac{4y-2y}{10x-4x}\)
\(A=\dfrac{5x}{-2x}+\dfrac{2y}{6x}=\dfrac{5}{-2}+\dfrac{1}{3}=\dfrac{-13}{6}\)