Bài 3: Giải hệ phương trình bằng phương pháp thế

\(\Delta'=m^2+2m+3=\left(m+\dfrac{1}{2}\right)^2+\dfrac{11}{4}>0\) với mọi m

Vậy pt luôn có 2 ngiệm phân biệt với mọi m

a. Bạn tự giải

b. Pt có nghiệm kép khi:

\(\Delta'=\left(m+1\right)^2-4m=0\Leftrightarrow m^2-2m+1=0\Leftrightarrow m=1\)

Khi đó: \(x_{1,2}=m+1=2\)

c. Do pt có nghiệm bằng 4:

\(\Rightarrow4^2-2\left(m+1\right).4+4m=0\)

\(\Leftrightarrow8-4m=0\Rightarrow m=2\)

\(x_1x_2=4m\Rightarrow x_2=\dfrac{4m}{x_1}=\dfrac{4.2}{4}=2\)

\(\Delta=4^2-4.1.(-1)=20>0\)

Theo Viét

\(\begin{cases}x_1+x_2=-4\\x_1x_2=1\end{cases}\)

\(A=\dfrac{x_1}{x_2}+\dfrac{x_2}{x_1}+\dfrac{5}{2}\)

\(=\dfrac{x_1^2+x_2^2}{x_1x_2}+\dfrac{5}{2}\)

\(=\dfrac{(x_1+x_2)^2-2x_1x_2}{x_1x_2}+\dfrac{5}{2}\)

\(=\dfrac{(-4)^2-2.1}{1}+\dfrac{5}{2}\)

\(=14+2,5=16,5\)

Vậy \(A=16,5\)

Giúp em với 😞

\(M=\dfrac{2x+y}{xy}+\dfrac{3}{2x+y}=\dfrac{2x+y}{2}+\dfrac{3}{2x+y}=\dfrac{3\left(2x+y\right)}{16}+\dfrac{3}{2x+y}+\dfrac{5}{16}\left(2x+y\right)\ge2\sqrt{\dfrac{3}{16}.3}+\dfrac{5}{16}.2\sqrt{2xy}=\dfrac{3}{2}+\dfrac{5}{4}=\dfrac{11}{4}\).

Đẳng thức xảy ra khi x = 1; y = 2.

\(M=\dfrac{2x+y}{xy}+\dfrac{3}{2x+y}=\dfrac{2x+y}{2}+\dfrac{3}{2x+y}\)

\(M=\dfrac{3\left(2x+y\right)}{16}+\dfrac{3}{2x+y}+\dfrac{5\left(2x+y\right)}{16}\ge2\sqrt{\dfrac{9\left(2x+y\right)}{16\left(2x+y\right)}}+\dfrac{5}{16}.2\sqrt{2xy}=\dfrac{11}{4}\)

Dấu "=" xảy ra khi \(\left(x;y\right)=\left(1;2\right)\)

Ta có: \(M=\dfrac{2x+y}{xy}+\dfrac{3}{2x+y}=\dfrac{2x+y}{2}+\dfrac{3}{2x+y}\)

\(=\left(\dfrac{3}{8}\dfrac{2x+y}{2}+\dfrac{3}{2x+y}\right)+\dfrac{5}{8}\dfrac{2x+y}{2}\)

Có: \(\dfrac{3}{8}\dfrac{2x+y}{2}+\dfrac{3}{2x+y}\ge2\sqrt{\dfrac{3}{8}\dfrac{2x+y}{2}\dfrac{3}{2x+y}}=\dfrac{3}{2}\)

Dấu '=' xảy ra <=> \(\dfrac{3}{8}\dfrac{2x+y}{2}=\dfrac{3}{2x+y}\)

Có: \(\dfrac{5}{8}\dfrac{2x+y}{2}\ge\dfrac{5}{8}\sqrt{2xy}=\dfrac{5}{4}\)

Dấu '=' xảy ra <=> 2x=y và xy=2

\(\Rightarrow M\ge\dfrac{3}{2}+\dfrac{5}{4}=\dfrac{11}{4}\)

Dấu '=' xảy ra <=> x=1, y=2

Vậy GTNN của M là 11/4 <=> x=1;y=2

`3x^4+2x^2=16`

Đặt `x^2=t (t >=0)` có:

`3t^2+2t=16`

`<=>3t^2+2t-16=0`

`<=>` \(\left[ \begin{array}{l}t=2(TM)\\t=\dfrac{-8}{3}(L)\end{array} \right.\) 

Với `t=2` : `x^2=2 <=> x= \pm \sqrt2`

Vậy `S={\pm \sqrt2}`.

Ta có: \(3x^4+2x^2=16\)

\(\Leftrightarrow3x^4+2x^2-16=0\)

\(\Leftrightarrow3x^4+6x^2-8x^2-16=0\)

\(\Leftrightarrow3x^2\left(x^2+2\right)-8\left(x^2+2\right)=0\)

\(\Leftrightarrow\left(x^2+2\right)\left(3x^2-8\right)=0\)

mà \(x^2+2>0\forall x\)

nên \(3x^2-8=0\)

\(\Leftrightarrow3x^2=8\)

\(\Leftrightarrow x^2=\dfrac{8}{3}\)

\(\Leftrightarrow x\in\left\{\dfrac{2\sqrt{6}}{3};-\dfrac{2\sqrt{6}}{3}\right\}\)

Vậy: \(S=\left\{\dfrac{2\sqrt{6}}{3};-\dfrac{2\sqrt{6}}{3}\right\}\)

`2x+5y=11(1)`

`2x-3y=0(2)`

Lấy (1) trừ (2)

`=>8y=11`

`<=>y=11/8`

`<=>x=(3y)/2=33/16`

a) Ta có: \(\left\{{}\begin{matrix}2x+5y=11\\2x-3y=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}8y=11\\2x-3y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{11}{8}\\2x=3y=3\cdot\dfrac{11}{8}=\dfrac{33}{8}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{33}{16}\\y=\dfrac{11}{8}\end{matrix}\right.\)

Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{33}{16}\\y=\dfrac{11}{8}\end{matrix}\right.\)

b) Ta có: \(\left\{{}\begin{matrix}4x+3y=6\\2x+y=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x+3y=6\\4x+2y=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-2\\2x+y=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-2=4\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=6\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-2\end{matrix}\right.\)

Vậy: Hệ phương trình có nghiệm duy nhất là (x,y)=(3;-2)

`b)4x+3y=6(1)`

`2x+y=4<=>4x+2y=8(2)`

Lấy (1) trừ (2) ta có:

`y=-2`

`<=>x=(4-2y)/2=3`