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

Help

2: \(ax+ay+bx+by\)

\(=a\left(x+y\right)+b\left(x+y\right)\)

\(=\left(x+y\right)\left(a+b\right)\)

3: \(x\left(x-2y\right)-x+2y\)

\(=x\left(x-2y\right)-\left(x-2y\right)\)

\(=\left(x-2y\right)\left(x-1\right)\)

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

Bài 3:

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

Sửa đề: Chứng minh 3a + 2 < 3b + 5

a ≤ b

⇒ 3a ≤ 3b

⇒ 3a + 2 ≤ 3b + 2 (1)

2 < 5

⇒ 3b + 2 < 3b + 5 (2)

Từ (1) và (2) ⇒ 3a + 2 < 3b + 5

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

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

?

câu nào

Câu 1:

1) Ta có: \(2x^2+5x-3=0\)

\(\Leftrightarrow2x^2+6x-x-3=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)

2) Để hàm số đồng biến trên R thì m-1>0

hay m>1

Câu 1: 

3) Ta có: P=a+b-2ab

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

\(=2-2\cdot\left(-1\right)=4\)

\(b,N=\left(2x-1\right)^2-4\ge-4\\ N_{min}=-4\Leftrightarrow x=\dfrac{1}{2}\\ c,P=\left(2x-5\right)^2+6\left(2x-5\right)+9-4\\ P=\left(2x-5+3\right)^2-4=\left(2x-2\right)^2-4\ge-4\\ P_{min}=-4\Leftrightarrow x=1\\ d,Q=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+1\\ Q=\left(x-1\right)^2+\left(y+2\right)^2+1\ge1\\ Q_{min}=1\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

6a.

$M=x^2-x+1=(x^2-x+\frac{1}{4})+\frac{3}{4}$

$=(x-\frac{1}{2})^2+\frac{3}{4}\geq \frac{3}{4}$

Vậy $M_{\min}=\frac{3}{4}$ khi $x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}$

Nãy ghi nhầm =="

a)Hđ gđ là nghiệm pt

`x^2=2x+2m+1`

`<=>x^2-2x-2m-1=0`

Thay `m=1` vào pt ta có:

`x^2-2x-2-1=0`

`<=>x^2-2x-3=0`

`a-b+c=0`

`=>x_1=-1,x_2=3`

`=>y_1=1,y_2=9`

`=>(-1,1),(3,9)`

Vậy tọa độ gđ (d) và (P) là `(-1,1)` và `(3,9)`

b)

Hđ gđ là nghiệm pt

`x^2=2x+2m+1`

`<=>x^2-2x-2m-1=0`

PT có 2 nghiệm pb

`<=>Delta'>0`

`<=>1+2m+1>0`

`<=>2m> -2`

`<=>m> 01`

Áp dụng hệ thức vi-ét:`x_1+x_2=2,x_1.x_2=-2m-1`

Theo `(P):y=x^2=>y_1=x_1^2,y_2=x_2^2`

`=>x_1^2+x_2^2=14`

`<=>(x_1+x_2)^2-2x_1.x_2=14`

`<=>4-2(-2m-1)=14`

`<=>4+2(2m+1)=14`

`<=>2(2m+1)=10`

`<=>2m+1=5`

`<=>2m=4`

`<=>m=2(tm)`

Vậy `m=2` thì ....