ĐKXĐ: -21\(\le x\le\)21

Đặt \(\left\{{}\begin{matrix}\sqrt{21+x}=a\\\sqrt{21-x}=b\end{matrix}\right.\left(a,b\ge0\right)\) (a\(\ne\)b)

Ta có \(\left\{{}\begin{matrix}21+x=a^2\\21-x=b^2\end{matrix}\right.\) =>\(\left\{{}\begin{matrix}a^2+b^2=42\\a^2-b^2=2x\end{matrix}\right.\)

Pt đã cho trở thành \(\dfrac{a+b}{a-b}=\dfrac{a^2+b^2}{a^2-b^2}\)

<=> \(\left(a+b\right)^2\)(a-b)=(\(a^2+b^2\))(a-b)

<=> (a-b)2ab=0

\(\text{​​}\text{​​}\left[{}\begin{matrix}a=b\left(loai\right)\\a=0\left(tm\right)\\b=0\left(tm\right)\end{matrix}\right.\)

Thay vào ta tìm dc S=\(\left\{21,-21\right\}\)

11/7 + 7/21 = 11 x 3 / 7 x 3 + 7/21 = 33/21 + 7/21 = 40/21

giúp mk câu nayvoiws

Chắc dưới mẫu bạn ghi nhầm căn đầu tiên

ĐKXĐ: \(-21\le x\le21;x\ne0\)

\(\Leftrightarrow\frac{\left(\sqrt{21+x}+\sqrt{21-x}\right)^2}{21+x-21+x}=\frac{21}{x}\)

\(\Leftrightarrow\frac{42+2\sqrt{21^2-x^2}}{2x}=\frac{21}{x}\)

\(\Leftrightarrow\sqrt{21^2-x^2}=0\)

\(\Rightarrow x=\pm21\)

bạn ơi, sao bước 2 làm thế nào mà đc bước 3 vậy ạ

21-21:21+21=21

21-21:21+21=21nha 
nhớ k

Chọn B

a: ĐKXĐ: -21<=x<=21 và x<>0

Ta có: \(\frac{\sqrt{21+x} + \sqrt{21-x}}{\sqrt{21+x} - \sqrt{21-x}} = \frac{21}{x}\)

=>\(\frac{(\sqrt{21+x} + \sqrt{21-x})^2}{(21+x) - (21-x)} = \frac{21}{x}\)

=>\(\frac{(21+x) + (21-x) + 2\sqrt{(21+x)(21-x)}}{2x} = \frac{21}{x}\)

=>\(\frac{42 + 2\sqrt{441 - x^2}}{2x} = \frac{21}{x}\)

=>\(\frac{21 + \sqrt{441 - x^2}}{x} = \frac{21}{x}\)

=>\(21+\sqrt{441-x^2}=21\)

=>\(\sqrt{441-x^2}=0\)

=>\(441-x^2=0\)

=>\(x^2=441\)

=>x=21(nhận) hoặc x=-21(nhận)

b: ĐKXĐ: x∈R

\(\left(\sqrt[3]{x+1}+\sqrt[3]{3x+1}\right)^3\)

\(=x+1+3x+1+3\cdot\sqrt[3]{\left(x+1\right)\left(3x+1\right)}\cdot\left(\sqrt[3]{x+1}+\sqrt[3]{3x+1}\right)\)

=4x+2+\(3\cdot\sqrt[3]{\left(x+1\right)\left(3x+1\right)}\cdot\left(\sqrt[3]{x+1}+\sqrt[3]{3x+1}\right)\)

Phương trình ban đầu sẽ trở thành:

\(4x + 2 + 3\sqrt[3]{(x+1)(3x+1)(x-1)} = x - 1\)

=>\(3\sqrt[3]{(x^2-1)(3x+1)}=-3x-3\)

=>\(\sqrt[3]{(x^2-1)(3x+1)}=-(x+1)\)

=>\((x^2-1)(3x+1) = -(x+1)^3\)

=>\((x-1)(x+1)(3x+1) + (x+1)^3 = 0\)

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

=>(x+1)(4x^2)=0

=>x=0 hoặc x=-1

Khi x=0 thì \(\sqrt[3]{1}+\sqrt[3]{1}=2<>\sqrt[3]{-1}=-1\) (loại)

Khi x=-1 thì \(\sqrt[3]{0}+\sqrt[3]{-2}=\sqrt[3]{-2}\) (nhận)

\(21-\left(2x-4\right)\left(x+1\right)\)

\(=21-2x^2-2x+4x+4\)

\(=-2x^2+2x+25\)