A. ( x -5 ) ( 7x + 1 ) - 7x ( x + 3)

= 7x² + x - 35x - 5 - 7x² - 21x

= (7x²-7x²) + (x - 35x - 21x) -5

= -56x - 5

B = (x² - 2x.2 + 2²) - x² + 1²

B = (x² - x²) - 4x + (2 + 1)

B= -4x +3

A. (x - 5)(7x + 1) - 7x(x + 3)

= 7x² + x - 35x - 5 - 7x² - 21x

= (7x² - 7x²) + (x - 35x - 21x) - 5

= -55x - 5

B. (x - 2)² - (x - 1)(x + 1)

= x² - 4x + 4 - x² + 1

= (x² - x²) - 4x + (4 + 1)

= -4x + 5

1: =>|x-5|=5-7x+7x+28=33

=>x-5=33 hoặc x-5=-33

=>x=38 hoặc x=-28

3: 2|x-6|+7x-2=|x-6|+7x

=>|x-6|=2

=>x-6=2 hoặc x-6=-2

=>x=8 hoặc x=4

a: x-4/5=3/7

=>x=3/7+4/5=43/35

b: x+3/7=4/5

=>x=4/5-3/7=13/35

c: 19/20-x=8/5-3/4

=>19/20-x=32/20-15/20=17/20

=>x=2/20=1/10

d: =>4/5*x=2/21

=>x=2/21:4/5=5/42

e: =>x:7/9=6/8

=>x=6/8*7/9=3/4*7/9=21/36=7/12

f: =>x/6=2/3-1/3=1/3

=>x=2

a: x-4/5=3/7

=>x=3/7+4/5=43/35

b: x+3/7=4/5

=>x=4/5-3/7=13/35

c: 19/20-x=8/5-3/4

=>19/20-x=32/20-15/20=17/20

=>x=2/20=1/10

d: =>4/5*x=11/21

=>x=55/84

e: =>x=6/8*7/9=42/72=7/12

f: =>x/6=2/3-1/3=1/3

=>x=2

1,\(K=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{x}}\)

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

\(=\dfrac{1}{\sqrt{2}}\left(\left|\sqrt{5}-1\right|+\sqrt{5}+1\right)\)\(=\dfrac{1}{\sqrt{2}}\left|\sqrt{5}-1+\sqrt{5}+1\right|=\dfrac{1}{\sqrt{2}}.2\sqrt{5}\)\(=\sqrt{10}\)

2, \(\sqrt{x-3}-2\sqrt{x^2-3x}=0\left(đk:x\ge3\right)\)

\(\Leftrightarrow\sqrt{x-3}\left(1-2\sqrt{x}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\1-2\sqrt{x}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=\left(\dfrac{1}{2}\right)^2=\dfrac{1}{4}\left(ktm\right)\end{matrix}\right.\)

Vậy pt có nghiệm x=3

3, \(\dfrac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\left(đk:x>-\dfrac{5}{7}\right)\)

\(\Leftrightarrow9x-7=7x+5\)

\(\Leftrightarrow x=6\left(tm\right)\)

4, \(x-5\sqrt{x}+4=0\)(đk: \(x\ge0\))

\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}=4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=16\end{matrix}\right.\) (tm)

Vậy...

1) Bạn tự làm

2) ĐK: \(x\ge3\)

PT \(\Leftrightarrow\sqrt{x-3}\left(1-2\sqrt{x}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\2\sqrt{x}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{4}\left(loại\right)\end{matrix}\right.\)

  Vậy ...

3) ĐK: \(x>-\dfrac{5}{7}\)

PT \(\Rightarrow9x-7=7x+5\) \(\Leftrightarrow x=6\)

  Vậy ...

4) ĐK: \(x\ge0\)

PT \(\Leftrightarrow x-4\sqrt{x}-\sqrt{x}+4=0\)

      \(\Leftrightarrow\left(\sqrt{x}-4\right)\left(\sqrt{x}-1\right)=0\)

      \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=4\\\sqrt{x}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=16\\x=1\end{matrix}\right.\)

  Vậy ...

\(7x.\left(x-2\right)-5.\left(x-1\right)=7x^2+3\)

\(7x^2-14x-5x+5=7x^2+3\)

\(7x^2-19x+5=7x^2+3\)

\(7x^2-19x+5-\left(7x^2+3\right)=0\)

\(7x^2-19x+5-7x^2-3=0\)

\(-19x+2=0\)

\(-19x=0-2\)

\(-19x=-2\)

\(x=\frac{2}{19}\)