cho mình xin tên sách

Bài 3:

\(b,\Leftrightarrow\left(x+8\right)\left(x+8-3x\right)=0\\ \Leftrightarrow\left(x+8\right)\left(8-2x\right)=0\\ \Leftrightarrow2\left(4-x\right)\left(x+8\right)=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-8\end{matrix}\right.\)

Bài 3: 

b: Ta có: \(\sqrt{x^2-2x+1}=\left|x-2\right|\)

\(\Leftrightarrow\left|x-1\right|=\left|x-2\right|\)

\(\Leftrightarrow x-1=2-x\)

\(\Leftrightarrow2x=3\)

hay \(x=\dfrac{3}{2}\)

Bài 4: ĐK: x>0

a)  \(B=\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+1-\dfrac{2x+\sqrt{x}}{\sqrt{x}}\)

\(\Leftrightarrow B=\dfrac{\sqrt{x}\left[\left(\sqrt{x}\right)^3+1\right]}{x-\sqrt{x}+1}+1-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}\)

\(\Leftrightarrow B=\dfrac{\sqrt{x}.\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+1-2\sqrt{x}-1\)

\(\Leftrightarrow B=\sqrt{x}.\left(\sqrt{x}+1\right)-2\sqrt{x}=x+\sqrt{x}-2\sqrt{x}\)

\(\Leftrightarrow B=x-\sqrt{x}\)

Vậy với x>0 thì \(B=x-\sqrt{x}\)

b) Ta có: \(B=2\)

\(\Leftrightarrow x-\sqrt{x}=2\)

\(\Leftrightarrow x-\sqrt{x}-2=0\)

 \(\Leftrightarrow x-2\sqrt{x}+\sqrt{x}-2=0\)

\(\Leftrightarrow\sqrt{x}.\left(\sqrt{x}-2\right)+\left(\sqrt{x}-2\right)=0\) 

\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)=0\)

Do \(\sqrt{x}+1>0\) nên, ta suy ra:

\(\sqrt{x}-2=0\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\) \(\left(TMĐK\right)\)

Vậy \(x=4\) thì \(B=2\)

Bài 4: 

a: Ta có: \(B=\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+1\)

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

\(=x+\sqrt{x}-2\sqrt{x}-1+1\)

\(=x-\sqrt{x}\)

b: Để B=2 thì \(x-\sqrt{x}-2=0\)

\(\Leftrightarrow\sqrt{x}-2=0\)

hay x=4

3b:

ĐKXĐ: \(x\in R\)

\(\sqrt{x^2+2x+4}+\left(x-1\right)\left(x+3\right)+1=0\)

=>\(\sqrt{x^2+2x+4}+x^2+2x-3+1=0\)

=>\(\sqrt{x^2+2x+4}+x^2+2x-2=0\)

=>\(x^2+2x+4+\sqrt{x^2+2x+4}-6=0\)

=>\(\left(\sqrt{x^2+2x+4}\right)^2+3\sqrt{x^2+2x+4}-2\sqrt{x^2+2x+4}-6=0\)

=>\(\left(\sqrt{x^2+2x+4}+3\right)\left(\sqrt{x^2+2x+4}-2\right)=0\)

=>\(\sqrt{x^2+2x+4}-2=0\)

=>\(\sqrt{x^2+2x+4}=2\)

=>\(x^2+2x+4=4\)

=>\(x^2+2x=0\)

=>x(x+2)=0

=>\(\left[{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-2\left(nhận\right)\end{matrix}\right.\)

Ptr hoành độ của `(P)` và `(d)` là:

       `x^2=-5x-3m+1`

`<=>x^2+5x+3m-1=0`   `(1)`

Để `(P)` cắt `(d)` tại `2` điểm phân biệt thì ptr `(1)` có `2` nghiệm phân biệt

    `=>\Delta > 0`

`<=>5^2-4(3m-1) > 0`

`<=>25-12m+4 > 0`

`<=>m < 29/12`

  `=>` Áp dụng Viét có: `{(x_1+x_2=-b/a=-5),(x_1.x_2=c/a=3m-1):}`

Ta có: `[x_1 ^2]/[x_2]-[x_2 ^2]/[x_1]+3=75/[x_1.x_2]`

`<=>[x_1 ^3-x_2 ^3]/[x_1.x_2]+[3x_1.x_2]/[x_1.x_2]=75/[x_1.x_2]`

  `=>(x_1-x_2)(x_1 ^2+x_1.x_2+x_2 ^2)+3x_1.x_2=75`

`<=>(x_1-x_2)[(x_1+x_2)^2-x_1.x_2]+3x_1.x_2=75`

`<=>(x_1-x_2)[(-5)^2-3m+1]+3(3m-1)=75`

`<=>(x_1-x_2)(26-3m)=78-9m`

`<=>x_1-x_2=[3(26-3m)]/[26-3m]`

`<=>x_1-x_2=3`

  Kết hợp với `x_1+x_2=-5`

Giải hệ  `=>{(x_1=-1),(x_2=-4):}`

Thay vào `x_1.x_2=3m-1` có:

    `-1.(-4)=3m-1`

`<=>m=5/3` (t/m)

b.

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}cos2x-\dfrac{1}{2}sin2x=-cosx\)

\(\Leftrightarrow cos\left(2x+\dfrac{\pi}{6}\right)=cos\left(x+\pi\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{6}=x+\pi+k2\pi\\2x+\dfrac{\pi}{6}=-x-\pi+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5\pi}{6}+k2\pi\\x=-\dfrac{7\pi}{18}+\dfrac{k2\pi}{3}\end{matrix}\right.\)

c.

\(\Leftrightarrow2cos4x.sin3x=2sin4x.cos4x\)

\(\Leftrightarrow cos4x\left(sin4x-sin3x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\sin4x=sin3x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{\pi}{2}+k\pi\\4x=3x+k2\pi\\4x=\pi-3x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\\x=k2\pi\\x=\dfrac{\pi}{7}+\dfrac{k2\pi}{7}\end{matrix}\right.\)

2.

\(f\left(x\right)=\dfrac{1}{2}-\dfrac{1}{2}cos2x-\dfrac{\sqrt{3}}{2}sin2x-5\)

\(=-\dfrac{9}{2}-\left(\dfrac{1}{2}cos2x+\dfrac{\sqrt{3}}{2}sin2x\right)\)

\(=-\dfrac{9}{2}-cos\left(2x-\dfrac{\pi}{3}\right)\)

Do \(-1\le-cos\left(2x-\dfrac{\pi}{3}\right)\le1\Rightarrow-\dfrac{11}{2}\le y\le-\dfrac{7}{2}\)

\(y_{min}=-\dfrac{11}{2}\) khi \(cos\left(2x-\dfrac{\pi}{3}\right)=1\Leftrightarrow x=\dfrac{\pi}{6}+k\pi\)

\(y_{max}=-\dfrac{7}{2}\) khi \(cos\left(2x-\dfrac{\pi}{3}\right)=-1\Rightarrow x=\dfrac{2\pi}{3}+k\pi\)

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