ai nay dung kinh nghiem la chinh

cau a)

ta thay \(10+6\sqrt{3}=\left(1+\sqrt{3}\right)^3\)

\(6+2\sqrt{5}=\left(1+\sqrt{5}\right)^2\)

khi do \(x=\frac{\sqrt[3]{\left(\sqrt{3}+1\right)^3}\left(\sqrt{3}-1\right)}{\sqrt{\left(1+\sqrt{5}\right)^2}-\sqrt{5}}\)

\(x=\frac{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}{1+\sqrt{5}-\sqrt{5}}\)

\(x=\frac{3-1}{1}=2\)

suy ra 

x^3-4x+1=1

A=1^2018

A=1

b)

ta thay

\(7+5\sqrt{2}=\left(1+\sqrt{2}\right)^3\)

khi do 

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

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

x=2

thay vao

x^3+3x-14=0

B=0^2018

B=0

bài 1:

a:\(\sqrt{\left(\sqrt{3}-2\right)^2}\)+\(\sqrt{\left(1+\sqrt{3}\right)^2}\)
=\(\sqrt{3}-2+1+\sqrt{3}\)
=\(2\sqrt{3}-1\)
b; dài quá mink lười làm thông cảm 
bài 2:
\(\sqrt{x^2-2x+1}=7\)
=>\(\sqrt{\left(x-1\right)^2}=7 \)
=>\(\orbr{\begin{cases}x-1=7\\x-1=-7\end{cases}}\)
=>\(\orbr{\begin{cases}x=8\\x=-6\end{cases}}\)
b: \(\sqrt{4x-20}-3\sqrt{\frac{x-5}{9}}=\sqrt{1-x}\)
=>\(\sqrt{4\left(x-5\right)}-9\sqrt{x-5}=\sqrt{1-x}\)
\(=2\sqrt{x-5}-9\sqrt{x-5}=\sqrt{1-x}\)
=>\(-7\sqrt{x-5}=\sqrt{1-x}\)
=\(-7.\left(x-5\right)=1-x\)
=>\(-7x+35=1-x\)
=>\(-7x+x=1-35\)
=>\(-6x=-34\)
=>\(x\approx5.667\)
mink sợ câu b bài 2 sai đó bạn

1 a)\(\sqrt{\left(\sqrt{3}-2\right)^2}+\sqrt{\left(1+\sqrt{3}\right)^2}\)

= \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1+\sqrt{3}\right)^2}\)

= \(|2-\sqrt{3}|+|1+\sqrt{3}|\)

= \(2-\sqrt{3}+1+\sqrt{3}\)

= \(2+1\)= \(3\)

b) \(\left(\frac{3}{2}\sqrt{6}+2\sqrt{\frac{2}{3}}-4\sqrt{\frac{3}{2}}\right)\cdot\left(3\sqrt{\frac{2}{3}}-\sqrt{12}-\sqrt{6}\right)\)

= \(\left(\frac{3}{2}\sqrt{6}+2\sqrt{\frac{6}{3^2}}-4\sqrt{\frac{6}{2^2}}\right)\cdot\left(3\sqrt{\frac{6}{3^2}}-\sqrt{6}\sqrt{2}-\sqrt{6}\right)\)

= \(\left(\frac{3}{2}\sqrt{6}+\frac{2}{3}\sqrt{6}-\frac{4}{2}\sqrt{6}\right)\cdot\left(\frac{3}{3}\sqrt{6}-\sqrt{6}\cdot\sqrt{2}-\sqrt{6}\right)\)

= \(\left(\frac{3}{2}\sqrt{6}+\frac{2}{3}\sqrt{6}-2\sqrt{6}\right)\cdot\left(\sqrt{6}-\sqrt{6}\cdot\sqrt{2}-\sqrt{6}\right)\)

= \(\left(\sqrt{6}\left(\frac{3}{2}+\frac{2}{3}-2\right)\right)\cdot\left(\sqrt{6}\left(1-\sqrt{2}-1\right)\right)\)

= \(\sqrt{6}\frac{1}{6}\cdot\sqrt{6}\left(-\sqrt{2}\right)\)

= \(\sqrt{6}^2\left(\frac{-\sqrt{2}}{6}\right)\)

= \(6\frac{-\sqrt{2}}{6}\)=\(-\sqrt{2}\)

2 a) \(\sqrt{x^2-2x+1}=7\)

<=> \(\sqrt{x^2-2x\cdot1+1^2}=7\)

<=> \(\sqrt{\left(x-1\right)^2}=7\)

<=> \(|x-1|=7\)

Nếu \(x-1>=0\)=>\(x>=1\)

=> \(|x-1|=x-1\)

\(x-1=7\)<=>\(x=8\)(thỏa)

Nếu \(x-1< 0\)=>\(x< 1\)

=> \(|x-1|=-\left(x-1\right)=1-x\)

\(1-x=7\)<=>\(-x=6\)<=> \(x=-6\)(thỏa)

Vậy x=8 hoặc x=-6

b) \(\sqrt{4x-20}-3\sqrt{\frac{x-5}{9}}=\sqrt{1-x}\)

<=> \(\sqrt{4\left(x-5\right)}-3\frac{\sqrt{x-5}}{3}=\sqrt{1-x}\)

<=> \(2\sqrt{x-5}-\sqrt{x-5}=\sqrt{1-x}\)

<=> \(\sqrt{x-5}=\sqrt{1-x}\)

ĐK \(x-5>=0\)<=> \(x=5\)

\(1-x\)<=> \(-x=-1\)<=> \(x=1\)

Ta có \(\sqrt{x-5}=\sqrt{1-x}\)

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

<=> \(x-5=1-x\)

<=> \(x-x=1+5\)

<=> \(0x=6\)(vô nghiệm)

Vậy phương trình vô nghiệm

Kết bạn với mình nha :)

a) ĐKXĐ: x\(\ge\)-3

PT\(\Leftrightarrow\sqrt{\left(x+7\right)\left(x+3\right)}=3\sqrt{x+3}+2\sqrt{x+7}-6\)

Đặt \(\left(\sqrt{x+3},\sqrt{x+7}\right)=\left(a,b\right)\)                 \(\left(a,b\ge0\right)\)

PT\(\Leftrightarrow ab=3a+2b-6\Leftrightarrow a\left(b-3\right)-2\left(b-3\right)=0\)

\(\Leftrightarrow\left(a-2\right)\left(b-3\right)=0\Leftrightarrow\orbr{\begin{cases}a=2\\b=3\end{cases}}\)(TM ĐK)

TH 1: a=2\(\Leftrightarrow\sqrt{x+3}=2\Leftrightarrow x+3=4\Leftrightarrow x=1\)(tm)

TH 2: b=3\(\Leftrightarrow\sqrt{x+7}=3\Leftrightarrow x+7=9\Leftrightarrow x=2\)(tm)

Vậy tập nghiệm phương trình S={1; 2}