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Ta có : M\(^2\)= (\(\dfrac{5x-4y}{5x+4y}\))\(^2\) = \(\dfrac{\left(5x-4y\right)^2}{\left(5x+4y\right)^2}\)= \(\dfrac{25x^2+16y^2-40xy}{25x^2+16y^2+40xy}\)
= \(\dfrac{41xy-40xy}{41xy+40xy}=\dfrac{xy}{81xy}=\dfrac{1}{81}=\left(\dfrac{1}{9}\right)^2\)
Mà 4y < 5x < 0 \(\Rightarrow\)5x - 4y > 0 . 5x +4y < 0 \(\Rightarrow\) M < 0
Vậy M = - \(\dfrac{1}{9}\)
a, \(ĐKXĐ:x\ne\pm\frac{1}{5},x\ne\frac{3}{2}\)
\(\Rightarrow P=\frac{\left(5x+1\right)\left(x+2\right)}{\left(2x-3\right)\left(5x-1\right)\left(5x+1\right)}-\frac{\left(8-3x\right)\left(5x+1\right)}{\left(5x-1\right)\left(5x+1\right)\left(2x-3\right)}\)
\(=\frac{x+2}{\left(2x-3\right)\left(5x-1\right)}-\frac{8-3x}{\left(5x-1\right)\left(2x-3\right)}\)
\(=\frac{2\left(2x-3\right)}{\left(2x-3\right)\left(5x-1\right)}=\frac{2}{5x-1}\)
b, Để P có giá trị nguyên thì \(2⋮5x-1\)
\(\Rightarrow5x-1\in\left\{1,2,-1,-2\right\}\)
=> x=..............
ĐKXĐ : x \(\ne\frac{3}{2}\) ; \(x\ne\frac{1}{5};x\ne-\frac{1}{5}\)
P= \(\frac{5x+1}{2x-3}.\left(\frac{x+2}{25x^2-1}-\frac{8-3x}{25x^2-1}\right)\)
P= \(\frac{5x-1}{2x-3}.\left(\frac{4x-6}{\left(5x+1\right).\left(5x-1\right)}\right)\)
P= \(\frac{5x-1}{2x-3}.\frac{2\left(2x-3\right)}{\left(5x-1\right)\left(5x+1\right)}\)
P= \(\frac{2}{5x-1}\)
KL
Em kiểm tra lại đề bài nhé vì:
\(Q=\left(x^3.x.y^n.y-\frac{1}{2}x^3.y^n.y^2\right):\frac{1}{2}x^3y^n-\left(4.5.x^2.x^2.y\right):\left(5x^2y\right)\)
\(=x^3y^n\left(xy-\frac{1}{2}y^2\right):\frac{1}{2}x^3y^n-5x^2y\left(4x^2\right):5x^2y\)
\(=2xy-y^2-4x^2=-\left(x^2-2xy+y^2\right)-3x^2=-\left[\left(x-y\right)^2+3x^2\right]< 0\)Với mọi x, y khác 0
=> Q luôn có gia trị âm với mọi x, y khác 0.
\(A=5x\left(4x^2-2x+1\right)-2x\left(10x^2-5x-2\right)\)
\(=20x^3-10x^2+5x-20x^3+10x^2+4x\)
\(=9x\)
Thay x=15 \(\Rightarrow A=9.15=135\)
\(B=6xy\left(xy-y^2\right)-8x^2\left(x-y^2\right)+5y^2\left(x^2-xy\right)\)
\(=6x^2y^2-6xy^3-8x^3+8x^2y^2+5x^2y^2-5xy^3\)
\(=19x^2y^2-11xy^3-8x^3\)
Thay x=1/2 ; y=2 vào B \(\Rightarrow19.\left(\frac{1}{2}\right)^2.2^2-11\cdot\frac{1}{2}\cdot2^3-8\cdot\left(\frac{1}{2}\right)^3\)
\(=19-44-1\)
\(=-26\)
\(25x^2+16y^2=50xy\)
\(\Leftrightarrow\) \(\left(5x+4y\right)^2-40xy=50xy\)
\(\Leftrightarrow\) \(\left(5x+4y\right)^2=90xy\)
Mặt khác, ta cũng có: \(25x^2+16y^2=50xy\)
\(\Leftrightarrow\) \(\left(5x-4y\right)^2=10xy\)
Do đó:
\(P^2=\frac{\left(5x-4y\right)^2}{\left(5x+4y\right)^2}=\frac{10xy}{90xy}=\frac{1}{9}\)
Vậy, \(P'=\frac{1+\frac{1}{9}}{1-\frac{1}{9}}=1\frac{1}{4}\)
1)
\(25x^2-40xy+16y^2=10xy\Leftrightarrow\left(5x-4y\right)^2=10xy\)
\(25x^2+40xy+16y^2=10xy\Leftrightarrow\left(5x+4y\right)^2=90xy\)
\(P^2=\frac{1}{9}\Leftrightarrow Q=\frac{1+P^2}{1-P^2}=\frac{1+\frac{1}{81}}{1-\frac{1}{81}}=\frac{82}{80}=\frac{41}{40}\)