b dễ làm trước,a ko biết làm   ):

b)\(\sqrt{2+\sqrt{x}}=3\)

ĐK : \(\sqrt{x}=7\)

\(x=49\)

\(\sqrt{2+\sqrt{49}}=3\Rightarrow\sqrt{2+7}=3\Leftrightarrow\sqrt{9}=3\Rightarrow3=3\)

\(\sqrt{\frac{1}{4}x^2+x+1}-\sqrt{6-2\sqrt{5}}=0\)

<=> \(\sqrt{\left(\frac{1}{2}x\right)^2+2\cdot\frac{1}{2}x\cdot1+1^2}-\sqrt{5-2\sqrt{5}+1}=0\)

<=> \(\sqrt{\left(\frac{1}{2}x+1\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}=0\)

<=> \(\left|\frac{1}{2}x+1\right|-\left|\sqrt{5}-1\right|=0\)

<=> \(\left|\frac{1}{2}x+1\right|-\left(\sqrt{5}-1\right)=0\)

<=> \(\left|\frac{1}{2}x+1\right|=\sqrt{5}-1\)

<=> \(\orbr{\begin{cases}\frac{1}{2}x+1=\sqrt{5}-1\\\frac{1}{2}x+1=1-\sqrt{5}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-4+2\sqrt{5}\\x=-2\sqrt{5}\end{cases}}\)

b) \(\sqrt{2+\sqrt{x}}=3\)

ĐK : x ≥ 0

Bình phương hai vế

pt <=> \(2+\sqrt{x}=9\)

    <=> \(\sqrt{x}=7\)

    <=> \(x=49\left(tm\right)\)

\(\text{Đ}k:a=b+c\)

\(min=2=1+1\)

\(\Rightarrow a=2,b=1,c=1\)

\(\frac{a^3+b^3}{a^3+c^3}=\frac{a+b}{a+c}\Rightarrow\frac{2^3+1^3}{2^3+1^3}=\frac{2+1}{2+1}\Leftrightarrow1=1\)

\(\Rightarrow\frac{a^3+b^3}{a^3+c^3}=\frac{a+b}{a+c}\)

Xét VT ta có :

\(VT=\frac{a^3+b^3}{a^3+c^3}=\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{\left(a+c\right)\left(a^2-ac+c^2\right)}\)

\(=\frac{\left(a+b\right)\left[\left(b+c\right)^2-\left(b+c\right)b+b^2\right]}{\left(a+c\right)\left[\left(b+c\right)^2-\left(b+c\right)c+c^2\right]}\)

\(=\frac{\left(a+b\right)\left(b^2+2bc+c^2-b^2-bc+b^2\right)}{\left(a+c\right)\left(b^2+2bc+c^2-bc-c^2+c^2\right)}\)

\(=\frac{\left(a+b\right)\left(b^2+bc+c^2\right)}{\left(a+c\right)\left(b^2+bc+c^2\right)}\)

\(=\frac{a+b}{a+c}=VP\)

=> đpcm

1. \(S=1+3+3^2+3^3+........+3^{2019}+3^{2020}\)

\(\Rightarrow3S=3+3^2+3^3+3^4+........+3^{2020}+3^{2021}\)

\(\Rightarrow3S-S=3^{2021}-1\)

\(\Rightarrow2S=3^{2021}-1\)

\(\Rightarrow S=\frac{3^{2021}-1}{2}\)

2. \(\left(3x-2\right)^3=64\)

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

\(\Leftrightarrow3x-2=4\)

\(\Leftrightarrow3x=6\)

\(\Leftrightarrow x=2\)

Vậy \(x=2\)

[3x-2]^3=64

Ta có:64=4^3    

Suy ra:3x-2=4

           3x   =4+2

          3x=6

         x=6:3

         x=2

\(12:\left\{390:\left[500-\left(5^3+7^2.5\right)\right]\right\}\)

\(=12:\left\{390:\left[500-\left(125+49.5\right)\right]\right\}\)

\(=12:\left\{390:\left[500-\left(125+245\right)\right]\right\}\)

\(=12:\left[390:\left(500-370\right)\right]\)

\(=12:\left(390:130\right)\)

\(=12:3\)

\(=4\)

   12 : { 390 : [ 500 - ( 5³  + 7²  . 5 ) ] }

= 12 : { 390 : [ 500 - (125 + 49 . 5 ) ] }

= 12 : { 390 : [ 500 - (125 + 245 ) ] }

= 12 : { 390 : [ 500 - 370] }

= 12 : { 390 : 130}

= 12 : 3

= 4

Xin mời XD

Các cặp góc so le trong là \(\widehat{A4}va\widehat{B2};\widehat{A3}va\widehat{B1}\)

Các cặp góc trong cùng phía là \(\widehat{A4}va\widehat{B1};\widehat{A3}va\widehat{B2}\)

Các cặp góc đồng vị là \(\widehat{A4}va\widehat{B4};\widehat{A3}va\widehat{B3};\widehat{A1}va\widehat{B1};\widehat{A2}va\widehat{B2}\)

Ta có \(\widehat{B4}=\widehat{B2}=40^o\) ( 2 góc đối đỉnh )

Ta có a//b 

\(\Rightarrow\widehat{A4}=\widehat{B2}=40^o\) (2 góc so le trong)

Lại có

\(\widehat{A2}=\widehat{B4}=40^o\) ( 2 góc đối đỉnh )

Vậy ...

Dòng thứ 2 từ dưới lên nhầm tí 

\(\widehat{A2}=\widehat{A4}=40^o\)(2 góc đối đỉnh )

sao ko ai trả lời vậy huhuhu

để mình trả lời