a, Áp dụng t/c dtsbn:

\(5x=7y\Rightarrow\dfrac{x}{7}=\dfrac{y}{5}=\dfrac{y-x}{5-7}=\dfrac{2}{-2}=-1\\ \Rightarrow\left\{{}\begin{matrix}x=-7\\y=-5\end{matrix}\right.\)

b, Áp dụng t/c dtsbn:

\(\dfrac{x}{y}=\dfrac{7}{2}\Rightarrow\dfrac{x}{7}=\dfrac{y}{2}=\dfrac{x+y}{7+2}=\dfrac{-27}{9}=-3\\ \Rightarrow\left\{{}\begin{matrix}x=-21\\y=-6\end{matrix}\right.\)

c, \(\dfrac{x}{32}=\dfrac{2}{x}\Rightarrow x^2=2\cdot32=64\Rightarrow\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\)

d, \(\left|x+\dfrac{1}{3}\right|-2=\dfrac{1}{2}\Rightarrow\left|x+\dfrac{1}{3}\right|=\dfrac{5}{2}\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{5}{2}\\x+\dfrac{1}{3}=-\dfrac{5}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{13}{6}\\x=-\dfrac{17}{6}\end{matrix}\right.\)

Lời giải:

a. $21-2(x+6)=15$

$2(x+6)=21-15=6$

$x+6=6:2=3$

$x=3-6=-3$

b.

$3^{x+1}=27=3^3$

$\Rightarrow x+1=3$

$\Rightarrow x=2$

c.

$14-x=36$

$x=14-36=-22$

d.

$-3x+25=-2$

$-3x=-2-25=-27$

$x=(-27):(-3)=9$

e.

$37-2(x+6)=29$

$2(x+6)=37-29=8$

$x+6=8:2=4$

$x=4-6=-2$

h.

$2^{x-1}=16=2^4$

$x-1=4$

$x=4+1=5$

a: 12:x=144

=>\(x=\dfrac{12}{144}\)

=>\(x=\dfrac{1}{12}\)

b: \(x-17=\left(-2\right)\cdot27\)

=>\(x-17=-54\)

=>\(x=-54+17=-37\)

c: \(3x-125=145\)

=>\(3x=125+145=270\)

=>\(x=\dfrac{270}{3}=90\)

\(a,12:x=144\\ x=\dfrac{12}{144}=\dfrac{1}{12}\\ ---\\ b,x-17=\left(-2\right).27\\ x-17=-54\\ x=-54+17=-37\\ ----\\ 3x-125=145\\ 3x=145+125=270\\ x=\dfrac{270}{3}=90\)

a: 12:x=144

x=\(\dfrac{1}{12}\)

b: 

c: 

x=90

a: \(=\dfrac{5\left(x+2\right)}{10xy^2}\cdot\dfrac{12x}{x+2}=\dfrac{60x}{10xy^2}=\dfrac{6}{y^2}\)

b: \(=\dfrac{x-4}{3x-1}\cdot\dfrac{3\left(3x-1\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{3}{x+4}\)

c: \(=\dfrac{2\left(2x+1\right)}{\left(x+4\right)^2}\cdot\dfrac{\left(x+4\right)}{3\left(x+3\right)}=\dfrac{2\left(2x+1\right)}{3\left(x+3\right)\left(x+4\right)}\)

d: \(=\dfrac{5\left(x-1\right)}{3\left(x+1\right)}\cdot\dfrac{x+1}{x-1}=\dfrac{5}{3}\)

a) (3x - 2)² - (1 + 5x)²

= (3x - 2 - 1 - 5x)(3x - 2 + 1 + 5x)

= (-2x - 3)(8x - 1)

b) (3x + 4)(3x - 4) - (5 - x)²

= (3x)² - 4² - (25 - 10x + x²)

= 9x² - 16 - 25 + 10x - x²

= 8x²  + 10x - 41

c) \(\left(\dfrac{1}{2}x+4\right)^2-\left(\dfrac{1}{2}x+3\right)\left(\dfrac{1}{2}x-3\right)\)

\(=\left(\dfrac{1}{2}x\right)^2+2.\dfrac{1}{2}x.4+4^2-\left[\left(\dfrac{1}{2}x\right)^2-3^2\right]\)

\(=\dfrac{1}{4}x^2+4x+16-\dfrac{1}{4}x^2+9\)

\(=4x+25\)

a: =9x^2-12x+4-25x^2-10x-1

=-16x^2-22x+3

b: =9x^2-16-x^2+10x-25

=8x^2+10x-41

c: \(=\dfrac{1}{4}x^2+4x+16-\dfrac{1}{4}x^2+9=4x+25\)

\(A\left(x\right)=5x^2-5x+3=5\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>0,\forall x\)

⇒ pt vô nghiệm

\(B\left(x\right)=4x^2-3x+7=4\left(x-\dfrac{3}{8}\right)^2+\dfrac{103}{16}>0,\forall x\)

⇒ pt vô nghiệm

\(C\left(x\right)=5x^2-11x+6=\left(5x^2-5x\right)-\left(6x-6\right)\)

\(=5x\left(x-1\right)-6\left(x-1\right)=\left(5x-6\right)\left(x-1\right)\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\\x=1\end{matrix}\right.\)

Vậy ...

a, Ta có : 

\(A\left(x\right)=5x^2-5x+1+2=0\Leftrightarrow5x^2-6x+3=0\)

\(\Leftrightarrow5\left(x^2-\dfrac{2.3}{5}+\dfrac{9}{25}-\dfrac{9}{25}\right)+3=0\Leftrightarrow5\left(x-\dfrac{3}{5}\right)^2+\dfrac{6}{5}=0\)( vô lí )

vậy đa thức ko có nghiệm 

b, \(B\left(x\right)=4x^2-3x+7=0\Leftrightarrow4\left(x^2-\dfrac{2.3}{8}+\dfrac{9}{64}-\dfrac{9}{64}\right)+7=0\)

\(\Leftrightarrow4\left(x-\dfrac{3}{8}\right)^2+\dfrac{103}{64}=0\)( vô lí ) 

Vậy đa thức ko có nghiệm 

c, \(C\left(x\right)=5x^2-11x+6=0\Leftrightarrow5x^2-6x-5x+6=0\)

\(\Leftrightarrow5x\left(x-1\right)-6\left(x-1\right)=0\Leftrightarrow\left(5x-6\right)\left(x-1\right)=0\Leftrightarrow x=\dfrac{6}{5};x=1\)