mk giải mới đc nửa bài thôi....

Bạn có cần gấp k?

mai mk trả lời nhé!

\(\Delta'=\left(m+1\right)^2-\left(m^2+4\right)=2m-3\ge0\Rightarrow m\ge\frac{3}{2}\) (1)

Khi đó theo Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1x_2=m^2+4\end{matrix}\right.\)

\(x_1^2+2\left(m+1\right)x_2\le3m^2+16\)

\(\Leftrightarrow x_1^2+\left(x_1+x_2\right)x_2\le3m^2+16\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-x_1x_2\le3m^2+16\)

\(\Leftrightarrow4\left(m+1\right)^2-\left(m^2+4\right)-3m^2-16\le0\)

\(\Rightarrow8m-16\le0\Rightarrow m\le2\) (2)

Kết hợp (1);(2) ta có: \(\frac{3}{2}\le m\le2\)

Không chắc đâu:v

a) Ta luôn có \(\left(x-1\right)^2+\left(2x-y-3\right)^2+\left(y+z\right)^2\ge0\forall x,y,z\)

Để đẳng thức xảy ra tức là \(\left(x-1\right)^2+\left(2x-y-3\right)^2+\left(y+z\right)^2=0\) (theo đề bài)

Thì \(\left\{{}\begin{matrix}x=1\\y=2x-3=2.1-3=-1\\z=-y=1\end{matrix}\right.\)

Vậy...

b) Ta luôn có \(VT\ge0\) với mọi x, y. Mà theo đề bài \(VT\le0\)

Do vậy \(VT=0\Leftrightarrow\left(2x+3\right)^{1998}+\left(3y-5\right)^{2000}=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-\frac{3}{2}\\y=\frac{5}{3}\end{matrix}\right.\)

`(x+1)(x+3)=2x^2-2`

`<=>x^2+x+3x+3=2x^2-2`

`<=>x^2-4x-5=0`

`<=>x^2-5x+x-5=0`

`<=>x(x-5)+(x-5)=0`

`<=>(x-5)(x+1)=0`

`<=>` $\left[ \begin{array}{l}x=5\\x=-1\end{array} \right.$

Vậy `S={5,-1}`

Ta có: \(\left(x+1\right)\left(x+3\right)=2x^2-2\)

\(\Leftrightarrow\left(x+1\right)\left(x+3\right)-2x^2+2=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+3\right)-2\left(x^2-1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+3\right)-2\left(x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left[x+3-2\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+3-2x+2\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(5-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\5-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)

Vậy: S={-3;5}

1.

ĐK: \(x\ne7;x\ne-1;x\ne3\)

\(\dfrac{2x-5}{x^2-6x-7}\le\dfrac{1}{x-3}\left(1\right)\)

TH1: \(x< -1\)

\(\left(1\right)\Leftrightarrow\left(2x-5\right)\left(x-3\right)\ge x^2-6x-7\)

\(\Leftrightarrow2x^2-11x+15\ge x^2-6x-7\)

\(\Leftrightarrow x^2-5x+22\ge0\)

\(\Leftrightarrow\) Bất phương trình đúng với mọi \(x< -1\)

TH2: \(-1< x< 3\)

\(\left(1\right)\Leftrightarrow\left(3-x\right)\left(2x-5\right)\ge\left(7-x\right)\left(x+1\right)\)

\(\Leftrightarrow-2x^2+11x-15\ge-x^2+6x+7\)

\(\Leftrightarrow-x^2+5x-22\ge0\)

\(\Rightarrow\) vô nghiệm

TH3: \(3< x< 7\)

Khi đó \(\dfrac{2x-5}{x^2-6x-7}\le0\); \(\dfrac{1}{x-3}>0\)

\(\Rightarrow\) Bất phương trình đúng với mọi \(3< x< 7\)

TH4: \(x>7\)

\(\left(1\right)\Leftrightarrow\left(2x-5\right)\left(x-3\right)\le x^2-6x-7\)

\(\Leftrightarrow2x^2-11x+15\le x^2-6x-7\)

\(\Leftrightarrow x^2-5x+22\le0\)

\(\Rightarrow\) vô nghiệm

Vậy ...

Các bài kia tương tự, chứ giải ra mệt lắm.

\(\frac{1}{2}\left(x-2\right)-2x\left(1-\frac{1}{2x}\right)=1\)

\(\Rightarrow\)\(\frac{1}{2}x-1-2x-1=1\)

\(\Rightarrow\)\(-\frac{3}{2}x-2=1\)

\(\Rightarrow\)\(-\frac{3}{2}x=2\)

\(\Rightarrow\)\(x=-\frac{4}{3}\)

2x^2-4x+x-2-[4x^2-8x+1]

=2x^2-3x-2-4x^2+8x-1

=-2x^2+5x-3

(2x + 1)(x - 2) - (2x - 1)²

= (2x + 1)(x - 2) - (4x² - 4x + 1)

= 2x² - 3x - 2x - 4x² + 4x - 1

= -2x² + x - 3

\(\left(2x+1\right)\left(x-2\right)-\left(2x-1\right)^2=2x^2-4x+x-2-4x^2+4x-1=-2x^2+x-3\)