đoạn cuối thiếu dấu"+"

\(A=\dfrac{\sqrt{4}-\sqrt{5}}{4-5}+\dfrac{\sqrt{5}-\sqrt{6}}{5-6}+....+\dfrac{\sqrt{34}-\sqrt{35}}{34-35}+\dfrac{\sqrt{35}-\sqrt{36}}{335-36}\)

\(A=\dfrac{\sqrt{4}-\sqrt{5}+\sqrt{5}-\sqrt{6}+....+\sqrt{35}-\sqrt{36}}{-1}=\dfrac{\sqrt{4}-\sqrt{36}}{-1}\)

\(A=\sqrt{36}-\sqrt{4}=6-2=4\)

mik cảm ơn ạ

Bài 2 

b, `\sqrt{3x^2}=x+2`          ĐKXĐ : `x>=0`

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

`=>3x^2=x^2+4x+4`

`=>3x^2-x^2-4x-4=0`

`=>2x^2-4x-4=0`

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

`=>(x^2-2x+1)-3=0`

`=>(x-1)^2=3`

`=>(x-1)^2=(\pm \sqrt{3})^2`

`=>` $\left[\begin{matrix} x-1=\sqrt{3}\\ x-1=-\sqrt{3}\end{matrix}\right.$

`=>` $\left[\begin{matrix} x=1+\sqrt{3}\\ x=1-\sqrt{3}\end{matrix}\right.$

Vậy `S={1+\sqrt{3};1-\sqrt{3}}`

mình nghĩ ĐKXĐ là như này : 

x+2≥0

➩ x≥-2

có phải k

\(b,\sqrt{2}.\sqrt{7+3\sqrt{5}}-\dfrac{4}{\sqrt{5}-1}\\ =\sqrt{14+6\sqrt{5}}-\dfrac{4}{\sqrt{5}-1}\\ =\sqrt{\sqrt{5^2}+2.3\sqrt{5}+3^2}-\dfrac{4}{\sqrt{5}-1}\\ =\sqrt{\left(\sqrt{5}+3\right)^2}-\dfrac{4}{\sqrt{5}-1}\\ =\left|\sqrt{5}+3\right|-\dfrac{4}{\sqrt{5}-1}\\ =\dfrac{\left(\sqrt{5}+3\right)\left(\sqrt{5}-1\right)-4}{\sqrt{5}-1}\\ =\dfrac{2+2\sqrt{5}-4}{\sqrt{5}-1}\\ =\dfrac{-2+2\sqrt{5}}{\sqrt{5}-1}\\ =\dfrac{2\left(-1+\sqrt{5}\right)}{\sqrt{5}-1}\\ =2\)

\(c,\sqrt{27}-6\sqrt{\dfrac{1}{3}}+\dfrac{\sqrt{3}-3}{\sqrt{3}}\\ =3\sqrt{3}-\dfrac{6}{\sqrt{3}}+\dfrac{\sqrt{3}-3}{\sqrt{3}}\)

\(=\dfrac{3\sqrt{3}.\sqrt{3}-6+\sqrt{3}-3}{\sqrt{3}}\\ =\dfrac{9-6+\sqrt{3}-3}{\sqrt{3}}\\ =\dfrac{\sqrt{3}}{\sqrt{3}}\\ =1\)

\(d,\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}\\ =\dfrac{\left(9-2\sqrt{3}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}{\left(3\sqrt{6}-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}\\ =\dfrac{27\sqrt{6}+18\sqrt{2}-18\sqrt{2}-4\sqrt{6}}{\left(3\sqrt{6}\right)^2-\left(2\sqrt{2}\right)^2}\\ =\dfrac{23\sqrt{6}}{54-8}\\ =\dfrac{23\sqrt{6}}{46}\\ =\dfrac{\sqrt{6}}{2}\)

Giải chi tiết từng bước nha

\(a,\)

\(=\left(\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right):\left(\dfrac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right)\)

\(=\left(\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right):\left(\dfrac{3}{3\sqrt{x}+1}\right)\)

\(=\dfrac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\dfrac{3\sqrt{x}+1}{3}\)

\(=\dfrac{3\sqrt{x}+3x}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\dfrac{3\sqrt{x}+1}{3}\)

\(=\dfrac{3\sqrt{x}\left(\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\dfrac{3\sqrt{x}+1}{3}\)

\(=\dfrac{3\sqrt{x}+1}{3\sqrt{x}-1}\)

Vậy \(P=\dfrac{3\sqrt{x}+1}{3\sqrt{x}-1}\)

\(b,\)Thay \(P=\dfrac{6}{5}\) vào pt, ta có :

\(\dfrac{3\sqrt{x}+1}{3\sqrt{x}-1}=\dfrac{6}{5}\)

\(\Leftrightarrow5\left(3\sqrt{x}+1\right)=6\left(3\sqrt{x}-1\right)\)

\(\Leftrightarrow15\sqrt{x}+5-18\sqrt{x}+6=0\)

\(\Leftrightarrow-3\sqrt{x}+11=0\)

\(\Leftrightarrow-3\sqrt{x}=-11\)

\(\Leftrightarrow\sqrt{x}=\dfrac{11}{3}\)

\(\Leftrightarrow x=\left(\dfrac{11}{3}\right)^2\)

\(\Leftrightarrow x=\dfrac{121}{9}\)

Vậy \(x=\dfrac{121}{9}\) thì \(P=\dfrac{6}{5}\)

b: Ta có: \(\sqrt[3]{-0.008}-\dfrac{1}{5}\cdot\sqrt[3]{64}+5\cdot\sqrt[3]{\left(-5\right)^3}\)

\(=-\dfrac{1}{5}-\dfrac{1}{5}\cdot4+5\cdot\left(-5\right)\)

\(=-\dfrac{1}{5}-\dfrac{4}{5}-25\)

=-26

đề bài là 0.08 mà bạn

\(a,ĐK:x\le\dfrac{5}{3}\\ PT\Leftrightarrow-3x+5=49\\ \Leftrightarrow x=-\dfrac{44}{3}\left(tm\right)\\ b,ĐK:x\ge-12\\ PT\Leftrightarrow\dfrac{1}{2}x+6=2\\ \Leftrightarrow\dfrac{1}{2}x=-4\\ \Leftrightarrow x=-8\left(tm\right)\\ c,ĐK:x\ge-\dfrac{1}{2}\\ PT\Leftrightarrow2x+1=13+4\sqrt{3}\\ \Leftrightarrow x=\dfrac{12+4\sqrt{3}}{2}=6+2\sqrt{3}\left(tm\right)\\ d,PT\Leftrightarrow\left|3x-1\right|=8\Leftrightarrow\left[{}\begin{matrix}3x-1=8\\1-3x=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{7}{3}\end{matrix}\right.\)

a) \(\sqrt{\dfrac{1}{8}}\cdot\sqrt{2}\cdot\sqrt{125}\cdot\sqrt{\dfrac{1}{5}}\) = \(\sqrt{\dfrac{1}{8}\cdot2}.\sqrt{125\cdot\dfrac{1}{5}}=\sqrt{\dfrac{1}{4}}.\sqrt{25}=\dfrac{1}{2}\cdot5=2,5\)

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

.

e) Ta có: \(\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)

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

=2