a: Khi x=2 thì (1) sẽ là:

4-2(m+2)+m+1=0

=>m+5-2m-4=0

=>1-m=0

=>m=1

x1+x2=m+1=3

=>x2=3-2=1

b: Δ=(m+2)^2-4(m+1)

=m^2+4m+4-4m-4=m^2>=0

=>Phương trình luôn có hai nghiệm

P=(x1+x2)^2-4x1x1+3x1x2

=(x1+x2)^2-x1x2

=(m+2)^2-m-1

=m^2+4m+4-m-1

=m^2+3m+3

=(m+3/2)^2+3/4>=3/4

Dấu = xảy ra khi m=-3/2

`(4\sqrt{6}+x)^2=8^2+(6+\sqrt{x^2+4})^2`

`<=>96+8\sqrt{6}x+x^2=64+36+12\sqrt{x^2+4}+x^2+4`

`<=>2\sqrt{6}x-2=3\sqrt{x^2+4}`    `ĐK: x >= \sqrt{6}/6`

`<=>24x^2-8\sqrt{6}x+4=9x^2+36`

`<=>15x^2-8\sqrt{6}x-32=0`

`<=>x^2-[8\sqrt{6}]/15x-32/15=0`

`<=>(x-[4\sqrt{6}]/15)^2-64/25=0`

`<=>|x-[4\sqrt{6}]/15|=8/5`

`<=>[(x=[24+4\sqrt{6}]/15 (t//m)),(x=[-24+4\sqrt{6}]/15(ko t//m)):}`

Bài nào em ??

Các bài ở hình ảnh trên

Bài 1: 

a) Ta có: \(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{2020\cdot2021}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2020}-\dfrac{1}{2021}\)

\(=1-\dfrac{1}{2021}\)

\(=\dfrac{2020}{2021}\)

b) Ta có: \(B=\dfrac{3}{3\cdot5}+\dfrac{3}{5\cdot7}+...+\dfrac{3}{47\cdot49}\)

\(=\dfrac{3}{2}\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{47\cdot49}\right)\)

\(=\dfrac{3}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{47}-\dfrac{1}{49}\right)\)

\(=\dfrac{3}{2}\left(\dfrac{1}{3}-\dfrac{1}{49}\right)\)

\(=\dfrac{3}{2}\cdot\dfrac{46}{147}\)

\(=\dfrac{46}{2}\cdot\dfrac{3}{147}\)

\(=\dfrac{23}{49}\)

Bài 2: 

Ta có: \(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}\)

\(=\dfrac{99}{100}< 1\)

Đề 1:

Bài 1:

\(a,=\sqrt{\left(\sqrt{7}+1\right)^2}-\left|-1+\sqrt{7}\right|=\sqrt{7}+1-\sqrt{7}+1=2\\ b,=2\sqrt{2}-4\sqrt{2}-5\sqrt{2}+\dfrac{\sqrt{2}}{2}=\dfrac{\sqrt{2}}{2}-7\sqrt{2}=\dfrac{-13\sqrt{2}}{\sqrt{2}}\)

Bài 2:

\(PT\Leftrightarrow\sqrt{\left(x-\dfrac{1}{2}\right)^2}=\dfrac{1}{2}\Leftrightarrow\left|x-\dfrac{1}{2}\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}+\dfrac{1}{2}=1\\x=-\dfrac{1}{2}+\dfrac{1}{2}=0\end{matrix}\right.\)

Bài 3:

\(a,M=\dfrac{a-2\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}=\dfrac{2\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)^2\left(\sqrt{a}+1\right)}=\dfrac{2}{\sqrt{a}+1}\\ b,M< 1\Leftrightarrow\dfrac{2}{\sqrt{a}+1}-1< 0\Leftrightarrow\dfrac{1-\sqrt{a}}{\sqrt{a}+1}< 0\\ \Leftrightarrow1-\sqrt{a}< 0\left(\sqrt{a}+1>0\right)\\ \Leftrightarrow a>1\)

phân số hả hay phép chia thường

\(\frac{4x}{6y}=\frac{2x+8}{3y+11}\Leftrightarrow\frac{2x}{3y}=\frac{2x+8}{3y+11}\Rightarrow2x\left(3y+11\right)=3y\left(2x+8\right)\Leftrightarrow6xy+11x=6yx+24y\)

\(\Leftrightarrow11x=24y\Leftrightarrow\frac{x}{y}=\frac{24}{11}\)