Cho x-2y=6. Tính
\(P=\dfrac{x-y}{x+6}\left(x\ne y;x\ne-6\right)\)
\(Q=\dfrac{2x+6}{3x-y}+\dfrac{2y-6}{4y-x}\left(3x\ne2y;4y\ne x\right)\)
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a)
\(\frac{x^2-16}{4x-x^2}=\frac{x^2-4^2}{x(4-x)}=\frac{(x-4)(x+4)}{x(4-x)}=\frac{x+4}{-x}\)
b) \(\frac{x^2+4x+3}{2x+6}=\frac{x^2+x+3x+3}{2(x+3)}=\frac{x(x+1)+3(x+1)}{2(x+3)}=\frac{(x+1)(x+3)}{2(x+3)}=\frac{x+1}{2}\)
c)
\(\frac{15x(x+y)^3}{5y(x+y)^2}=\frac{5.3.x(x+y)^2.(x+y)}{5y(x+y)^2}=\frac{3x(x+y)}{y}\)
d) \(\frac{5(x-y)-3(y-x)}{10(x-y)}=\frac{5(x-y)+3(x-y)}{10(x-y)}=\frac{8(x-y)}{10(x-y)}=\frac{8}{10}=\frac{4}{5}\)
e) \(\frac{2x+2y+5x+5y}{2x+2y-5x-5y}=\frac{7x+7y}{-3x-3y}=\frac{7(x+y)}{-3(x+y)}=\frac{-7}{3}\)
f) \(\frac{x^2-xy}{3xy-3y^2}=\frac{x(x-y)}{3y(x-y)}=\frac{x}{3y}\)
g) \(\frac{2ax^2-4ax+2a}{5b-5bx^2}=\frac{2a(x^2-2x+1)}{5b(1-x^2)}=\frac{2a(x-1)^2}{5b(1-x)(1+x)}\)
\(=\frac{2a(x-1)}{5b(-1)(x+1)}=\frac{2a(1-x)}{5b(x+1)}\)
Đề bài sai, đề đúng thì phân thức đằng sau dấu chia phải là:
\(\dfrac{4x^4+4x^2y+y^2-4}{x^2+y+xy+x}\)
a: \(A=\dfrac{6}{7}x^2y^2\cdot\dfrac{-7}{2}x^2y=-3x^4y^3\)
b: Áp dụng tính chất của dãy tỉ số bằng nhau,ta được:
\(\dfrac{x}{-2}=\dfrac{y}{3}=\dfrac{x-y}{-2-3}=\dfrac{5}{-5}=-1\)
Do đó: x=2; y=-3
\(A=-3x^4y^3=-3\cdot2^4\cdot\left(-3\right)^3=3\cdot27\cdot16=81\cdot16=1296\)
\(A=\dfrac{6}{7}x^2y^2.\left(-3\dfrac{1}{2}x^2y\right)\)
\(=\dfrac{6}{7}x^2y^2.\left(-\dfrac{7}{2}\right)x^2y\)
\(=-3x^4y^3\)
b)Có: \(\dfrac{x}{y}=-\dfrac{2}{3}\Leftrightarrow\dfrac{x}{2}=\dfrac{-y}{3}=\dfrac{x-y}{2+3}=\dfrac{5}{5}=1\)
\(\Rightarrow x=2;y=-3\)
Tại \(x=2;y=-3\) , giá trị của biểu thức là:
\(-3.2^4.\left(-3\right)^3=-3.16.\left(-27\right)=1296\)
a: =-4xyz^2
b: =-9x^2y
c: =16x^2y^2
d: =1/6x^2y^3
e: =13/6x^3y^2
f: =7/12x^4y
a) -xyz² - 3xz.yz
= -xyz² - 3xyz²
= -4xyz²
b) -8x²y - x.(xy)
= -8x²y - x²y
= -9x²y
c) 4xy².x - (-12x²y²)
= 4x²y² + 12x²y²
= 16x²y²
d) 1/2 x²y³ - 1/3 x²y.y²
= 1/2 x²y³ - 1/3 x²y³
= 1/6 x²y³
e) 3xy(x²y) - 5/6 x³y²
= 3x³y² - 5/6 x³y²
= 13/6 x³y²
f) 3/4 x⁴y - 1/6 xy.x³
= 3/4 x⁴y - 1/6 x⁴y
= 7/12 x⁴y
Lời giải:
Đặt $\frac{x-1}{x+2y}=a; \frac{y+1}{x-2y}=b$ thì HPT trở thành:
\(\left\{\begin{matrix}
5a+3b=8\\
20a-7b=-6\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}
20a+12b=32\\
20a-7b=-6\end{matrix}\right.\)
\(\Rightarrow 19b=38\Rightarrow b=2\Rightarrow a=0,4\)
Ta có:
\(\left\{\begin{matrix} a=\frac{2}{5}\\ b=2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} \frac{x-1}{x+2y}=\frac{2}{5}\\ \frac{y+1}{x-2y}=2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 3x=4y+5\\ 2x=1+5y\end{matrix}\right.\)
\(\Rightarrow 2(4y+5)-3(1+5y)=0\Rightarrow y=1\)
Kéo theo $x=3$
Vậy $(x,y)=(3,1)$
Lời giải:
Ta có:
\(\frac{4x^2y^2}{(x^2+y^2)^2}+\frac{x^2}{y^2}+\frac{y^2}{x^2}\geq 3\)
\(\Leftrightarrow \frac{4x^2y^2}{(x^2+y^2)^2}-1+\frac{x^2}{y^2}+\frac{y^2}{x^2}-2\geq 0\)
\(\Leftrightarrow \frac{4x^2y^2-(x^2+y^2)^2}{(x^2+y^2)^2}+\left(\frac{x}{y}-\frac{y}{x}\right)^2\geq 0\)
\(\Leftrightarrow \frac{-(x^2-y^2)^2}{(x^2+y^2)^2}+\frac{(x^2-y^2)^2}{x^2y^2}\geq 0\)
\(\Leftrightarrow (x^2-y^2)^2\left(\frac{1}{x^2y^2}-\frac{1}{(x^2+y^2)^2}\right)\geq 0\)
\(\Leftrightarrow \frac{(x^2-y^2)^2(x^4+y^4+x^2y^2)}{x^2y^2(x^2+y^2)^2}\geq 0\) (luôn đúng)
Do đó ta có đpcm.
Dấu bằng xảy ra khi $x=y$
\(A=\dfrac{4x^2y^2}{\left(x^2+y^2\right)^2}+\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\)
x,y khác 0
<=>\(A=\dfrac{4}{\left(\dfrac{x}{y}+\dfrac{y}{x}\right)^2}+\left(\dfrac{x}{y}\right)^2+\left(\dfrac{y}{x}\right)^2\)
\(A+2=\dfrac{4}{\left(\dfrac{x}{y}+\dfrac{y}{x}\right)^2}+\left(\dfrac{x}{y}+\dfrac{y}{x}\right)^2=m\)
\(\left(\dfrac{x}{y}+\dfrac{y}{x}\right)^2=t;t\ge4\)
\(m=\dfrac{4}{t}+t\Leftrightarrow t^2-mt+4=0\)
f(t) có nghiệm t>= 4<=>\(\left\{{}\begin{matrix}m^2-16\ge0\\\dfrac{m+\sqrt{m^2-16}}{2}\ge4\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left|m\right|\ge4\\m^2-16\ge m^2-16m+64\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left|m\right|\ge4\\m\ge5\end{matrix}\right.\) \(\Leftrightarrow A+2\ge5;A\ge3=>dpcm\)
x-2y=6
nên x=2y+6
\(P=\dfrac{2y+6-y}{2y+6+6}=\dfrac{y+6}{2y+12}=\dfrac{1}{2}\)
\(Q=\dfrac{2x+6}{3x-2y}+\dfrac{2y-6}{4y-x}\)
\(=\dfrac{2\left(2y+6\right)}{3\left(2y+6\right)-2y}+\dfrac{2y-6}{4y-\left(2y+6\right)}\)
\(=\dfrac{4y+12}{6y+18-2y}+\dfrac{2y-6}{4y-2y-6}\)
\(=\dfrac{4y+12}{4y+18}+\dfrac{2y-6}{2y-6}\)
\(=\dfrac{8y^2-24y+24y-48+8y^2+36y-24y-108}{\left(4y+18\right)\left(2y-6\right)}\)
\(=\dfrac{16y^2+12y-156}{4\left(2y+9\right)\cdot\left(y-3\right)}\)
\(=\dfrac{4y^2+3y-34}{\left(2y+9\right)\left(y-3\right)}\)