a.

\(\sqrt{x^2-4x+1}=x\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2-4x+1=x^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\-4x+1=0\end{matrix}\right.\)

\(\Rightarrow x=\dfrac{1}{4}\)

b.

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

\(\Leftrightarrow\left\{{}\begin{matrix}x+1\ge0\\5x^2-2x+2=\left(x+1\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\4x^2-4x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x=\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow x=\dfrac{1}{2}\)

c.

\(\sqrt{x^2-8x+16}=4-x\)

\(\Leftrightarrow\sqrt{\left(4-x\right)^2}=4-x\)

\(\Leftrightarrow\left|4-x\right|=4-x\)

\(\Leftrightarrow4-x\ge0\)

\(\Rightarrow x\le4\)

d.

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

\(\Leftrightarrow\left\{{}\begin{matrix}4x-3\ge0\\3x+1=4x-3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\x=4\end{matrix}\right.\)

\(\Rightarrow x=4\)

\(\dfrac{28}{36}=\dfrac{28\div4}{36\div4}=\dfrac{7}{9}\)

\(\dfrac{21}{81}=\dfrac{21\div3}{81\div3}=\dfrac{7}{27}\)

\(\dfrac{150}{100}=\dfrac{150\div50}{100\div50}=\dfrac{3}{2}\)

\(\dfrac{70}{80}=\dfrac{70\div10}{80\div10}=\dfrac{7}{8}\)

\(15)15+4\left(x-2\right)=95\\ 4\left(x-2\right)=95-15\\ 4\left(x-2\right)=80\\ x-2=\dfrac{80}{4}\\ x-2=20\\ x=20+2\\ x=22\\ 16)20-\left(x+14\right)=5\\ x+14=20-5\\ x+14=15\\ x=15-14\\ x=1\\ 17)24+3\left(5-x\right)=27\\ 3\left(5-x\right)=27-24\\ 3\left(5-x\right)=3\\ 5-x=\dfrac{3}{3}=1\\ x=5-1=4\\ 18)15:x=5\\ x=15:5\\ x=3\\ 19)\dfrac{x}{4}=3\\ x=4\cdot3\\ x=12\\ 20)\dfrac{21}{x}=7\\ x=\dfrac{21}{7}\\ x=3\)

\(A=10\cdot\dfrac{4}{7}-1,38-7,62\\ =10\cdot\dfrac{4}{7}-\left(1,38+7,62\right)\\ =\dfrac{10\cdot4}{7}-9\\ =\dfrac{40}{7}-\dfrac{63}{7}\\ =\dfrac{40-63}{7}\\ =\dfrac{-23}{7}\)

\(1)A=x^2-7x+2\\ =\left(x^2-2\cdot x\cdot\dfrac{7}{2}+\dfrac{49}{4}\right)-\dfrac{41}{4}\\ =\left(x-\dfrac{7}{2}\right)^2-\dfrac{41}{4}\)

Ta có: `(x-7/2)^2>=0` với mọi x

`=>A=(x-7/2)^2-41/4>=-41/4` với mọi x

Dấu "=" xảy ra: `x-7/2=0<=>x=7/2` 

\(2)B=9x^2-12x+5\\ =\left(9x^2-12x+4\right)+1\\ =\left[\left(3x\right)^2-2\cdot3x\cdot2+2^2\right]+1\\ =\left(3x-2\right)^2+1\)

Ta có: `(3x-2)^2>=0` với mọi x

`=>B=(3x-2)^2+1>=1` với mọi x

Dấu "=" xảy ra: `3x-2=0<=>x=2/3` 

`23*16+23*84-300`

`=23*(16+84)-300`

`=23*100-300`

`=2300-300`

`=2000 `

3) 

\(a,3+\dfrac{1}{2}\\ =\dfrac{6}{2}+\dfrac{1}{2}\\ =\dfrac{6+1}{2}\\ =\dfrac{7}{2}\\ b,\dfrac{5}{8}+2\\ =\dfrac{5}{8}+\dfrac{16}{8}\\ =\dfrac{5+16}{8}\\ =\dfrac{21}{8}\\ c,3-\dfrac{5}{7}\\ =\dfrac{21}{7}-\dfrac{5}{7}\\ =\dfrac{21-5}{7}\\ =\dfrac{16}{7}\\ d,\dfrac{35}{12}-2\\ =\dfrac{35}{12}-\dfrac{24}{12}\\ =\dfrac{35-24}{12}\\ =\dfrac{11}{12}\)

2 nha các bạn

\(36500g=\dfrac{36500}{3600}h\)

Gọi d là ƯCLN của `2n+3` và `4n+7`

Ta có: 

`2n+3` ⋮ d và `4n+7` ⋮ d 

`=>2(2n+3)` ⋮ d và `4n+7` ⋮ d

`=>4n+6` ⋮ d và `4n+7` ⋮ d 

`=>(4n+7)-(4n+6)` ⋮ d 

`=>1` ⋮ d 

`=>d=1` 

Vậy: `2n+3` và `4n+7` là 2 số nguyên tố cùng nhau 

Mọi người giúp em với ạ, bây giờ em đang cần gấp ạ!