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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 \(F=\frac{4y+2\cdot\frac{4y}{3}}{5y-7\cdot\frac{4y}{3}}+\frac{3\cdot\frac{4y}{3}-2y}{10y-4\cdot\frac{4y}{3}}=\frac{4y+\frac{8y}{3}}{5y-\frac{28y}{3}}+\frac{4y-2y}{10y-\frac{16y}{3}}=\frac{-20}{13}+\frac{3}{7}=\frac{-101}{91}\)
2:
a: \(=\left(2x^2-xy\right)+\left(2xz-yz\right)\)
\(=x\left(2x-y\right)+z\left(x-2y\right)=\left(x-2y\right)\left(x+z\right)\)
b: \(=\left(x^2-4y^2\right)-\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x+2y-1\right)\)
c: \(=\left(y^2+10y+25\right)-9z^2\)
\(=\left(y+5\right)^2-\left(3z\right)^2\)
\(=\left(y+5+3z\right)\left(y+5-3z\right)\)
d: \(=\left(x+2y\right)^3-\left(x-2y\right)\left(x+2y\right)\)
\(=\left(x+2y\right)\left[\left(x+2y\right)^2-\left(x-2y\right)\right]\)
\(=\left(x+2y\right)\left(x^2+4xy+4y^2-x+2y\right)\)
1:
a: \(x\left(3-4x\right)+5\left(3-4x\right)=\left(3-4x\right)\left(x+5\right)\)
b: \(2y\left(5y-6\right)-4\left(6-5y\right)\)
\(=2y\left(5y-6\right)+4\left(5y-6\right)\)
\(=2\left(5y-6\right)\left(y+2\right)\)
c: \(=27\left(x-2\right)^3-3x\left(x-2\right)^2\)
\(=3\left(x-2\right)^2\cdot\left[9\left(x-2\right)-x\right]\)
\(=3\left(x-2\right)^2\left(8x-18\right)=6\left(x-2\right)^2\cdot\left(4x-9\right)\)
d: \(=6y\left(x-y\right)\left(x+y\right)-8y\left(x+y\right)^2\)
\(=2y\left(x+y\right)\left[3\left(x-y\right)-4\left(x+y\right)\right]\)
\(=2y\left(x+y\right)\left(3x-3y-4x-4y\right)\)
\(=2y\left(x+y\right)\left(-x-7y\right)\)
Bài 1
a) x(3 - 4x) + 5(3 - 4x)
= (3 - 4x)(x + 5)
b) 2y(5y - 6) - 4(6- 5y)
= 2y(5y - 6) + 4(5y - 6)
= (5y - 6)(2y + 4)
= 2(5y - 6)(y + 2)
c) 27(x - 2)³ - 3x(2 - x)²
= 27(x - 2)³ - 3x(x - 2)²
= 3(x - 2)²[9(x - 2) - x]
= 3(x - 2)²(9x - 18 - x)
= 3(x - 2)²(8x - 18)
= 6(x - 2)²(4x - 9)
d) 6y(x² - y²) - 8y(x + y)²
= 6y(x - y)(x + y) - 8y(x + y)²
= 2y(x + y)[3(x - y) - 4(x + y)]
= 2y(x + y)(3x - 3y - 4x - 4y)
= 2y(x + y)(-x - 7y)
= -2y(x + y)(x + 7y)
Ta có: \(A^2=\dfrac{\left(3x-2y\right)^2}{\left(3x+2y\right)^2}\)
\(=\dfrac{9x^2+4x^2-12xy}{9x^2+4x^2+12xy}\)
\(=\dfrac{20xy-12xy}{20x^2+12xy}\)
\(=\dfrac{8xy}{32xy}=\dfrac{1}{4}\)
\(\Leftrightarrow A\in\left\{\dfrac{1}{2};-\dfrac{1}{2}\right\}\)(1)
Vì 2y<3x<0 nên 3x-2y>0 và 3x+2y<0
hay \(A=\dfrac{3x-2y}{3x+2y}< 0\)(2)
Từ (1) và (2) suy ra \(A=-\dfrac{1}{2}\)
Vậy: \(A=-\dfrac{1}{2}\)
\(A=-4x^2-5y^2+8xy+10y+12\)
\(-A=4x^2+5y^2-8xy-10y-12\)
\(-A=\left(4x^2-8xy+y^2\right)+\left(4y^2-10y+\frac{25}{4}\right)-\frac{73}{4}\)
\(-A=\left(2x-y\right)^2+\left(2y-\frac{5}{2}\right)^2-\frac{73}{4}\)
Mà : \(\left(2x-y\right)^2\ge0\forall x;y\)
\(\left(2y-\frac{5}{2}\right)^2\ge0\forall y\)
\(\Rightarrow-A\ge-\frac{73}{4}\)
\(\Leftrightarrow A\le\frac{73}{4}\)
Dấu "=" xảy ra khi :
\(\hept{\begin{cases}2x-y=0\\2y-\frac{5}{2}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{5}{8}\\y=\frac{5}{4}\end{cases}}\)
Vậy \(A_{Max}=\frac{73}{4}\Leftrightarrow\left(x;y\right)=\left(\frac{5}{8};\frac{5}{4}\right)\)
f: \(=\dfrac{5x-3-x+3}{4x^2y}=\dfrac{4x}{4x^2y}=\dfrac{1}{xy}\)
g: \(=\dfrac{3x+10-x-4}{x+3}=\dfrac{2x+6}{x+3}=2\)
h: \(=\dfrac{4-2+x}{x-1}=\dfrac{x+2}{x-1}\)
n: \(=\dfrac{3x-x+6}{x\left(x+3\right)}=\dfrac{2\left(x+3\right)}{x\left(x+3\right)}=\dfrac{2}{x}\)
p: \(=\dfrac{x^2-9-x^2+9}{x\left(x-3\right)}=0\)
k: \(=\dfrac{x-2x-4+x-2}{\left(x+2\right)\left(x-2\right)}=\dfrac{-6}{x^2-4}\)
m: \(=\dfrac{3x-x+6}{x\left(2x+6\right)}=\dfrac{2x+6}{x\left(2x+6\right)}=\dfrac{1}{x}\)
bài 1:
a) x(x-2)-5y-(x-2)=(x-5y)(x-2)
b) =(2x-3-4x)(2x-3+4x)=(-2x-3)(6x-3)
bài 2 bạn tự luyện nhé
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!!!
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:
\(A=\dfrac{4y+2x}{5y-7x}+\dfrac{3x-2y}{10y-4x}\)
\(A=\dfrac{3x+2x}{5x-7x}+\dfrac{4y-2y}{10x-4x}\)
\(A=\dfrac{5x}{-2x}+\dfrac{2y}{6x}=\dfrac{5}{-2}+\dfrac{1}{3}=\dfrac{-13}{6}\)