a)\(-5\sqrt{80}+4\sqrt{45}-2\sqrt{245}\)

\(=-20\sqrt{5}+12\sqrt{5}-14\sqrt{5}\)

\(=\left(-20+12-14\right)\sqrt{5}=-22\sqrt{5}\)

b)\(\sqrt{12-6\sqrt{3}}-\sqrt{21-12\sqrt{3}}\)

\(=\sqrt{9-6\sqrt{3}+3}-\sqrt{12-6\sqrt{12}+9}\)

\(=\sqrt{\left(3-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{12}-3\right)^2}\)

\(=\left|3-\sqrt{3}\right|-\left|\sqrt{12}-3\right|\)

\(=3-\sqrt{3}-\sqrt{12}+3\)(do \(3>\sqrt{3};\sqrt{12}>3\))

\(=6-\sqrt{12}-\sqrt{3}\)

\(=6-2\sqrt{3}-\sqrt{3}=6-3\sqrt{3}\)

c)\(\sqrt{7-\sqrt{40}}-\sqrt{7+\sqrt{40}}\)

\(=\sqrt{5-2\sqrt{5}.\sqrt{2}+2}-\sqrt{5+2\sqrt{5}.\sqrt{2}+2}\)

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

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

\(=\sqrt{5}-\sqrt{2}-\sqrt{5}-\sqrt{2}\)(do \(\sqrt{5}>\sqrt{2}\))

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

ĐK: \(x< 0\)hoặc \(x>\frac{1}{3}\).

Đặt \(t=\sqrt{\frac{3x-1}{x}}>0\)

Phương trình ban đầu tương đương với:

\(2t=\frac{1}{t^2}+1\)

\(\Leftrightarrow2t^3-t^2-1=0\)

\(\Leftrightarrow2t^3-2t^2+t^2-1=0\)

\(\Leftrightarrow\left(t-1\right)\left(2t^2+t+1\right)=0\)

\(\Leftrightarrow t-1=0\)(vì \(2t^2+t+1>0\))

\(\Leftrightarrow t=1\)

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

\(\Rightarrow3x-1=x\)

\(\Leftrightarrow x=\frac{1}{2}\)(thỏa mãn) 

Mình mới thử chương trình lớp 9 nên chưa hiểu nhiều lắm. Cảm ơn nhé!

\(\sqrt{12-3\sqrt{15}}=\frac{1}{\sqrt{2}}\sqrt{24-6\sqrt{15}}=\frac{1}{\sqrt{2}}\sqrt{15-2.3.\sqrt{15}+9}\)

\(=\frac{1}{\sqrt{2}}\sqrt{\left(\sqrt{15}-3\right)^2}=\frac{1}{\sqrt{2}}\left|\sqrt{15}-3\right|=\frac{\sqrt{15}-3}{\sqrt{2}}\)

\(=\frac{\sqrt{30}-3\sqrt{2}}{2}\)

mình điền sai ở câu a) phải là 4x nhé

Mình viết lại câu a) \(\sqrt{2x^2+4x+5}\)

\(A=\frac{2\sqrt{x}+1+2\sqrt{x}-1-5+\sqrt{x}}{x-1}\)

\(A=\frac{5\sqrt{x}-5}{x-1}\)

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

\(A=\frac{5}{\sqrt{x}+1}\)

x= -1 nha bạn

\(\sqrt{x^2-8x+16}=5\)
\(\Leftrightarrow\sqrt{\left(x-4\right)^2}=5\)
\(\Leftrightarrow\left|x-4\right|=5\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=5\\x-4=-5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=9\\x=-1\end{cases}}\)

Bài 1:

a,\(\left(\frac{1}{2}x+4\right)^2=\frac{1}{4}x^2+4x+16\)

b,(7x-5y)²=49x²-70xy+25y²

c,(6x²+y²)(y²-6x²)=(y²+6x²)(y²-6x²)=y⁴+36x⁴

Bài 2 :

a,(5x+1)³=125x³+75x²+15x+1

b,(x-2y)³=x³-6x²y+12xy²-8y³

c,(4x+5)(16x²-20x+25)=64x³+125

d,(6x-\(\frac{1}{3}\))(36x²+2x+\(\frac{1}{9}\))=216x³-\(\frac{1}{27}\)

Bài 3 :

a,(2x+3)²+(2x-3)²-2(4x²-9)

=4x²+12x+9+4x²-12x+9-8x²+18

=36

b,(x+2)³+(x-2)³+x³-3x(x+2)(x-2)

= x³+6x²+12x+8+x³-6x²+12x-8+x³-3x³+12x

=36x

Bài 4 : 

A=(3x+2)²+(2x-7)²-2(3x+2)(2x-7)

A=9x²+12x+4+4x²-28x+49-12x²+42x-8x+28

A=x²+18x+81

A=(x+9)²

Thay x=-19 vào biểu thức

=> A=(-19+9)²

     A=(-10)²=100

Bài 5 :

B=(3x-1)²-(x+7)²-2(2x-5)(2x+5)

B= 9x²-6x+1-x²+14x-49-8x²+50

B=8x+2

B=2(4x+1)

Thay x=\(\frac{1}{5}\)vào biểu thức

=> B=2(4.\(\frac{1}{5}\)+1)

      B=2.\(\frac{9}{5}\)=\(\frac{18}{5}\)