Do \(2\le a\le6\Rightarrow\sqrt{a-2}\ge0;\sqrt{a-2}+2\ge0;\sqrt{a-2}-2\le0\)\(\sqrt{a+4\sqrt{a-2}+2}+\sqrt{a-4\sqrt{a-2}+2}=\sqrt{\left(a-2\right)+4\sqrt{a-2}+4}+\sqrt{\left(a-2\right)-4\sqrt{a-2}+4}=\sqrt{\left(\sqrt{a-2}+2\right)^2}+\sqrt{\left(\sqrt{a-2}-2\right)^2}=\left|\sqrt{a-2}+2\right|+\left|\sqrt{a-2}-2\right|=\sqrt{a-2}+2-\left(\sqrt{a-2}-2\right)=4\left(đpcm\right)\)

\(\sqrt{a-2+4\sqrt{a-2}+4}+\sqrt{a-2-4\sqrt{a-2}+4}\)

\(\Leftrightarrow\sqrt{\left(\sqrt{a-2}+2\right)^2}+\sqrt{\left(\sqrt{a-2}-2\right)^2}\)

\(\Leftrightarrow\left|\sqrt{a-2}+2\right|+\left|\sqrt{a-2}-2\right|\)

Theo BĐT Cosi ta có \(\left|\sqrt{a-2}+2\right|+\left|\sqrt{a-2}-2\right|\ge\left|\sqrt{a-2}+2+2-\sqrt{a-2}\right|=4\)

Dấu ''='' xảy ra khi 2 =< a =< 6 

\(\left(\sqrt{\dfrac{8}{3}}-\sqrt{24}+\sqrt{\dfrac{50}{3}}\right).\sqrt{6}\)

=\(\left(\dfrac{\sqrt{8}}{\sqrt{3}}-\dfrac{6\sqrt{2}}{\sqrt{2}}+\dfrac{\sqrt{50}}{\sqrt{3}}\right).\sqrt{6}\)

=\(\dfrac{\sqrt{6}}{3}.\sqrt{6}\)

=\(\dfrac{6}{3}=2\)

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a, y = -x + 2b - 5 (d') 

(d') đi qua A(-3;4) <=> 4 = 3 + 2b - 5 <=> 2b = 6 <=> b = 3 

b, Cho điểm giao giữa (d) ; (d') là A(3a;a) với a là tung độ 

(d) đi qua A(3a;a) <=> a = 6a - 3 <=> -5a = -3 <=> a = 3/5 

=> A(9/5;3/5) đi qua (d') \(\dfrac{3}{5}=-\dfrac{9}{5}+2b-5\Leftrightarrow\dfrac{12}{5}+5=2b\Leftrightarrow b=\dfrac{37}{10}\)

Hình bạn tự vẽ nhé

a) Dễ dàng tính được \(BC=HB+HC=2+6=8\left(cm\right)\)

Tam giác ABC vuông tại A, đường cao AH nên \(AB^2=BH.BC\left(htl\right)\Rightarrow AB=\sqrt{BH.BC}=\sqrt{2.8}=4\left(cm\right)\)

Tương tự, ta có \(AC^2=CH.BC\Rightarrow AC=\sqrt{CH.BC}=\sqrt{6.8}=4\sqrt{3}\left(cm\right)\)

Mặt khác theo htl:\(AH.BC=AB.AC\Rightarrow AH=\dfrac{AB.AC}{BC}=\dfrac{4.4\sqrt{3}}{8}=2\sqrt{3}\left(cm\right)\)

b) Tam giác ABH vuông tại H có đường cao HD nên \(AH^2=AD.AB\left(htl\right)\)

Tương tự, ta có \(AH^2=AE.AC\), từ đó \(AD.AB=AE.AC\left(=AH^2\right)\)  (đpcm)

c) Ta có \(AD.AB=AE.AC\left(cmt\right)\Rightarrow\dfrac{AE}{AB}=\dfrac{AD}{AC}\)

Xét \(\Delta AED\) và \(\Delta ABC\), ta có \(\dfrac{AE}{AB}=\dfrac{AD}{AC}\left(cmt\right);\) \(\widehat{A}\) chung

\(\Rightarrow\Delta AED~\Delta ABC\left(c.g.c\right)\) \(\Rightarrow\widehat{ADE}=\widehat{ACB}\) hay \(\widehat{ADE}=\widehat{BCE}\)

Mà \(\widehat{ADE}+\widehat{BDE}=180^o\) \(\Rightarrowđpcm\)

Dùng cách phân tích thành nhân tử, ta có thể viết phương trình như sau :

\(x^3\left(x-4\right)-19x\left(x-4\right)+30\left(x-4\right)=0\)

hay \(\left(x-4\right)\left(x^3-19x+30\right)=0\)

      \(\left(x-4\right)\left(x^3-3x^2-9x-10x+30\right)=0\)

Ta được phương trình tích sau : \(\left(x-4\right)\left(x-3\right)\left(x^2-3x-10\right)=0\)

\(S=x_1=4;x_2=3;x_i=2;x_4=-5\)

\(\dfrac{x}{1}+\dfrac{x}{1+2}+\dfrac{x}{1+2+3}+...+\dfrac{x}{1+2+3+...+4041}=4041\)

<=> \(x\left(1+\dfrac{1}{\dfrac{2.3}{2}}+\dfrac{1}{\dfrac{3.4}{2}}+...+\dfrac{1}{\dfrac{4041.4042}{2}}\right)=4041\)

<=> \(2x\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{4041.4042}\right)=4041\)

<=> \(2x\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{4041}-\dfrac{1}{4042}\right)=4041\)

<=> \(2x\left(1-\dfrac{1}{4042}\right)=4041\)

<=> \(\dfrac{4041x}{2021}=4041\Leftrightarrow x=2021\)

\(=x\times\left(1+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+3+...+4041}\right)=4041\Leftrightarrow x\times\left(\dfrac{2}{2}+\dfrac{2}{2\times3}+\dfrac{2}{3\times4}+...+\dfrac{2}{4041\times4042}\right)=4041\Leftrightarrow2x\times\left(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{4041\times4042}\right)=4041\Leftrightarrow x\times(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{4041}-\dfrac{1}{4042})=\dfrac{4041}{2}\Leftrightarrow x\times\left(1-\dfrac{1}{4042}\right)=\dfrac{4041}{2}\Leftrightarrow x\times\dfrac{4041}{4042}=\dfrac{4041}{2}\Leftrightarrow x=\dfrac{4041}{2}:\dfrac{4041}{4042}\Leftrightarrow x=2021\)Vậy x = 2021

Em mới lớp 8 nên ko bt có chỗ nào sai hay ko mong anh/chị bỏ qua

`[sin \alpha + cos \alpha-1]/[1-cos \alpha]=[2cos \alpha]/[sin \alpha-cos \alpha+1]`

`<=>(sin \alpha+cos \alpha-1)(sin \alpha-cos \alpha+1)=2cos \alpha(1-cos \alpha)`

`<=>sin^2 \alpha-(cos \alpha-1)^2=2cos \alpha-2cos^2 \alpha`

`<=>sin^2 \alpha-cos^2 \alpha+2cos \alpha-1=2cos \alpha-2cos^2 \alpha`

`<=>sin^2 \alpha+cos^2 \alpha=1` (LĐ)

Vậy đẳng thức đc c/m

`\sqrt{x+12+6\sqrt{x+3}}-\sqrt{x+12-6\sqrt{x+3}}`   `ĐK: x >= -3`

`=\sqrt{(\sqrt{x+3})^2+2.\sqrt{x+2}.3+3^2}-\sqrt{(\sqrt{x+3})^2-2.\sqrt{x+2}.3+3^2}`

`=\sqrt{(\sqrt{x+3}+3)^2}-\sqrt{(\sqrt{x+3}-3)^2}`

`=|\sqrt{x+3}+3|-|\sqrt{x+3}-3|`

`=\sqrt{x+3}+3-|\sqrt{x+3}-3|`

`@` Với `\sqrt{x+3}-3 >= 0<=>\sqrt{x+3} >= 3<=>x+3 >= 9<=>x >= 6` (t/m)

    `=>\sqrt{x+3}+3-|\sqrt{x+3}-3|=\sqrt{x+3}+3-\sqrt{x+3}+3=6`

`@` Với `\sqrt{x+3}-3 < 0<=>\sqrt{x+3} < 3<=>x+3 < 9<=>x < 6`

                                    Kết hợp đk `x >= -3 =>-3 <= x < 6`

   `=>\sqrt{x+3}+3-|\sqrt{x+3}-3|=\sqrt{x+3}+3-3+\sqrt{x+3}=2\sqrt{x+3}`

\(\sqrt{x+12+6\sqrt{x+3}}-\sqrt{x+12-6\sqrt{x+3}}\) \(\left(ĐKXĐ:x\ge-3\right)\)

\(=\sqrt{\left(x+3\right)+2\sqrt{x+3}.3+9}-\sqrt{\left(x+3\right)-2\sqrt{x+3}.3+9}\)

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

\(=|\left(\sqrt{x}+3\right)+3|-|\left(\sqrt{x}+3\right)-3|\)

\(=\left(\sqrt{x}+3\right)+3-\left(\sqrt{x}+3\right)+3=6\) ( Với \(x\ge-3\) )