a)\(\left(\sqrt{21}+7\right)\cdot\sqrt{10-2\sqrt{21}}\)

\(=\left(\sqrt{21}+7\right)\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)

\(=\sqrt{7}\left(\sqrt{3}+\sqrt{7}\right)\left(\sqrt{7}-\sqrt{3}\right)\)

\(=\sqrt{7}\left(7-3\right)=4\sqrt{7}\)

b)\(\left(7+\sqrt{14}\right)\sqrt{9-2\sqrt{14}}\)

\(=\left(7+\sqrt{14}\right)\sqrt{\left(\sqrt{7}-\sqrt{2}\right)^2}\)

\(=\sqrt{7}\left(\sqrt{7}+\sqrt{2}\right)\left(\sqrt{7}-\sqrt{2}\right)\)

\(=\sqrt{7}\left(7-2\right)=5\sqrt{7}\)

giup minh voi minh can gap lam 

Ta có:

\(\left(\frac{2}{11.13}+\frac{2}{13.15}+...+\frac{2}{19.21}\right).231-\left(3^{16}:3^{14}-2^8:2^6\right).x=2013^0\)

\(\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{19}-\frac{1}{21}\right).231-\left(3^2-2^2\right).x=1\)

\(\left(\frac{1}{11}-\frac{1}{21}\right).231-x=1\)

\(\frac{10}{231}.231-x=1\)

\(10-x=1\)

\(x=9\)

Vậy,.............

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b: \(=\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)\sqrt{10-2\sqrt{21}}\)

\(=\left(5+\sqrt{21}\right)\left(10-2\sqrt{21}\right)\)

\(=50-10\sqrt{21}+10\sqrt{21}-42=8\)

a: \(A=\sqrt{\sqrt{2}-1}+\sqrt{\sqrt{2}+1}\)

=>\(A^2=\sqrt{2}-1+\sqrt{2}+1+2\sqrt{2-1}=2\sqrt{2}+2\)

=>\(A=\sqrt{2\sqrt{2}+2}\)

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

=>\(B=\sqrt{2\sqrt{2}+2}-\sqrt{2+\sqrt{2}}\)

=>\(B^2=2\sqrt{2}+2+2+\sqrt{2}-2\sqrt{\sqrt{2}\left(2+\sqrt{2}\right)\left(2+\sqrt{2}\right)}\)

=>\(B^2=4+3\sqrt{2}-2\sqrt[4]{2}\left(2+\sqrt{2}\right)\)

=>\(B\simeq0,35\)

b: Phương trình cần tìm là x^2+8x-105=0

=>(x+15)(x-7)=0

=>x=-15 hoặc x=7

c: Phương trình có hai nghiệm u,v thỏa mãn là x^2-2x+9=0

=>PTVN

d: Phương trình có hai nghiệm u,v thỏa mãn là x^2-5x+24=0

=>PTVN

trả lời giúp em

ĐKXĐ: \(x\ge-2;y\ge-11\)

\(x\left(x+21\right)+y\left(x-33\right)=2\left(y^2+50\right)\)

\(\Leftrightarrow x^2+\left(y+21\right)x-2y^2-33y-100=0\)

\(\Delta=\left(y+21\right)^2+4\left(2y^2+33y+100\right)=\left(3y+29\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-y-21+3y+29}{2}=y+4\\x=\dfrac{-y-21-3y-29}{2}=-2y-25\end{matrix}\right.\)

TH1: \(x=-2y-25\Rightarrow x+2y=-25\)

Mà \(x+2y\ge-2+2.\left(-11\right)=-23>-25\)

\(\Rightarrow\) Pt vô nghiệm

TH2:  \(x=y+4\) thay vào pt dưới:

\(\sqrt{y+6}+2\sqrt{y+11}=\sqrt{\left(3y+10\right)^3}\)

\(\Leftrightarrow\sqrt{y+6}-2+2\sqrt{y+11}-6=\sqrt{\left(3y+10\right)^3}-8\)

\(\Leftrightarrow\dfrac{y+2}{\sqrt{y+6}+2}+\dfrac{2\left(y+2\right)}{\sqrt{y+11}+3}=\dfrac{3\left(y+2\right)\left(3y+14+2\sqrt{3y+10}+4\right)}{\sqrt{3y+10}+2}\)

\(\Leftrightarrow\left[{}\begin{matrix}y=-2\Rightarrow x=2\\\dfrac{1}{\sqrt{y+6}+2}+\dfrac{2}{\sqrt{y+11}+3}=\dfrac{3\left(3y+14+2\sqrt{3y+10}+4\right)}{\sqrt{3y+10}+2}\left(1\right)\end{matrix}\right.\)

Xét (1), ta có:

\(\dfrac{1}{\sqrt{y+6}+2}+\dfrac{2}{\sqrt{y+11}+3}< \dfrac{1}{2}+\dfrac{2}{3}< 2\)

\(\dfrac{3\left(3y+14+2\sqrt{3y+10}+4\right)}{\sqrt{3y+10}+2}=\dfrac{3\left(3y+14\right)}{\sqrt{3y+10}+2}+6>2\)

\(\Rightarrow\left(1\right)\) vô nghiệm

Vậy hệ có nghiệm duy nhất \(\left(x;y\right)=\left(2;-2\right)\)

https://hoc24.vn/cau-hoi/.5005119341955 hỗ trợ em với thầy

\(\left|5x-3\right|-2x=14\\ \Leftrightarrow\left|5x-3\right|=14+2x\\ \Leftrightarrow\left[{}\begin{matrix}5x-3=14+2x\\5x-3=-14-2x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=17\\7x=-11\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{3}\\x=-\dfrac{11}{7}\end{matrix}\right.\)

https://olm.vn/hoi-dap/detail/41801581362.html

Xem giùm nhé 

=6.23-6.6/6.22+6.12

=6.(23-6)/6.(22+12)

=6.17/6.34

=6.17/6.2.17

=1/2

=2.7.(- 2).(- 11).(- 2).(- 5).(- 2)/(- 11).2.2.(- 5).7.3

= 2.(- 2)/7.3

=- 4/21

n) \(\left|3-x\right|+x^2-x\left(x+4\right)=0\)

\(\Rightarrow\left|3-x\right|+x^2-x^2-4x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3-x-4x=0\left(đk:3-x\ge0\right)\\-\left(3-x\right)-4x=0\left(đk:3-x< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\left(đk:x\le3\right)\\x=-1\left(đk:x>3\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x\in\varnothing\end{matrix}\right.\)

Vậy \(x=\dfrac{3}{5}\)

m) \(\left(x-1\right)^2+\left|x+21\right|-x^2-13=0\)

\(\Rightarrow x^2-2x+1+\left|x+21\right|-x^2-13=0\)

\(\Leftrightarrow-2x-12+\left|x+21\right|=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x-12+x+21=0\left(đk:x+21\ge0\right)\\-2x-12-\left(x+21\right)=0\left(đk:x+21< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=9\left(đk:x\ge-21\right)\\x=-11\left(đk:x< -21\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=9\\x\in\varnothing\end{matrix}\right.\)

Vậy \(x=9\)

e) \(\left|5x\right|=3x-2\)

\(\Rightarrow5\cdot\left|x\right|=3x-2\)

\(\Leftrightarrow5\cdot\left|x\right|-3x=-2\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-3x=-2\left(đk:x\ge0\right)\\5\cdot\left(-x\right)-3x=-2\left(đk:x< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(đk:x\ge0\right)\\x=\dfrac{1}{4}\left(đk:x< 0\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x\in\varnothing\\x\in\varnothing\end{matrix}\right.\)

Vậy \(x\in\varnothing\)

g) \(\left|-2,5x\right|=x-12\)

\(\Rightarrow2,5\cdot\left|x\right|=x-12\)

\(\Leftrightarrow2x5\cdot\left|x\right|-x=-12\)

\(\Leftrightarrow\left[{}\begin{matrix}2,5x-x=-12\left(đk:x\ge0\right)\\2,5\cdot\left(-x\right)-x=-12\left(đk:x< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-8\left(đk:x\ge0\right)\\x=\dfrac{24}{7}\left(đk:x< 0\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x\in\varnothing\\x\in\varnothing\end{matrix}\right.\)

Vậy \(x\in\varnothing\)