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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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: \(A=-\dfrac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)Trừ theo vế của \(pt\left(2\right)\) cho \(pt\left(1\right)\):
\(\left(5x+3y\right)-\left(3x+2y\right)=-4-1\)
\(\Leftrightarrow2x+y=-5\). Khi đó
\(3x+2y=1\Leftrightarrow2\left(2x+y\right)-x=1\)
\(\Leftrightarrow2\cdot\left(-5\right)-x=1\)\(\Leftrightarrow x=-11\)
\(\Rightarrow3x+2y=1\Rightarrow y=\dfrac{1-3x}{2}=\dfrac{1-3\cdot\left(-11\right)}{2}=17\)
Vậy nghiệm hpt \(\left(x;y\right)=\left(-11;17\right)\)
b)\(2x^2+2\sqrt{3}x-3=0\)
\(\Delta=\left(2\sqrt{3}\right)^2-\left(4\cdot2\cdot\left(-3\right)\right)=36\)
\(\Rightarrow x_{1,2}=\dfrac{-2\sqrt{3}\pm\sqrt{36}}{4}\)
c)\(9x^4+8x^2-1=0\)
\(\Leftrightarrow9x^4-x^2+9x^2-1=0\)
\(\Leftrightarrow x^2\left(9x^2-1\right)+\left(9x^2-1\right)=0\)
\(\Leftrightarrow\left(9x^2-1\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)\left(x^2+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3x-1=0\\3x+1=0\\x^2+1=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\pm\dfrac{1}{3}\\x^2+1>0\left(loai\right)\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Ta có:
\(\left\{\begin{matrix} x+2y+3z=4\\ \frac{1}{x}+\frac{1}{2y}+\frac{1}{3z}=0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x+2y+3z=4\\ \frac{6yz+2xy+3xz}{6xyz}=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x+2y+3z=4\\ 2xy+6yz+3xz=0\end{matrix}\right.\)
Do đó:
\((x+2y+3z)^2-2(2xy+6yz+3xz)=4^2-2.0=16\)
\(\Leftrightarrow x^2+4y^2+9z^2=16\)
\(\Leftrightarrow P=16\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 :
\(A=\dfrac{2y-y}{2y+y}=\dfrac{y}{3y}=\dfrac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{3x-2y}{3x+2y}=\dfrac{\left(3x-2y\right)^2}{\left(3x+2y\right)^2}=\dfrac{9x^2+4y^2-12xy}{9x^2+4y^2+12xy}=\dfrac{1}{4}\)
thay từ đề vào ok
![](https://rs.olm.vn/images/avt/0.png?1311)
M=\(\dfrac{1}{2}\)
nha bạn
liệu có cần giải hẳn ra không nhỉ?
ta có: 9x^2+4y^2=20xy=> 9x^2-2.2.3xy+4y^2=8xy
=> (3x-2y)^2=8xy
mặt khác 9x^2+4y^2=20xy=> 9x^2+2.2.3xy+4y^2=32xy
=>(3x+2y)^2=32xy
=>(3x-2y)^2/(3x+2y)^2=8xy/32xy=1/4
=>(3x-2y)/(3x+2y)=căn 1/4=1/2 hoặc -1/2
mà x<2y=>x=-1/2
Ta có:
\(9x^2+4y^2=20xy\)
\(\Leftrightarrow9x^2-20xy+4y^2=0\)
\(\Leftrightarrow9x^2-18xy-2xy+4y^2=0\)
\(\Leftrightarrow9x\left(x-2y\right)-2y\left(x-2y\right)=0\)
\(\Leftrightarrow\left(x-2y\right)\left(9x-2y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2y\\9x=2y\end{matrix}\right.\)
Mà \(x< 2y\) nên \(9x=2y\Leftrightarrow x=\dfrac{2}{9}y\) (1)
Thay (1) vào A ta được:
\(A=\dfrac{3.\dfrac{2}{9}y-2y}{3.\dfrac{2}{9}y+2y}=\dfrac{y\left(\dfrac{2}{3}-2\right)}{y\left(\dfrac{2}{3}+2\right)}=\dfrac{-\dfrac{4}{3}}{\dfrac{8}{3}}=-\dfrac{1}{2}\)
Vậy..................................