a) Ta có: f(x)=-3

<=>x⁵-2x²+x⁴-x⁵+3x²-x⁴-3+2x=-3

<=>(x⁵-x⁵)+(-2x²+3x²)+(x⁴-x⁴)+2x-3=-3

<=>x²+2x-3=-3

<=>x²+2x=0

<=>x(x+2)=0

<=>x=0 hoặc x+2=0

<=>x=0 hoặc x=-2

Vậy..........

b)đa thức f(x) có nghiệm

<=>f(x)=0

<=>x²+2x-3=0

<=>x²+3x-x-3=0

<=>x(x+3)-(x+3)=0

<=>(x-1)(x+3)=0

<=>x-1=0 hoặc x+3=0

<=>x=1 hoặc x=-3

Vậy nghiệm của đa thức f(x) là x=-3;x=1

a, f(x)=0

b, x=1 và x=-3

a) \(2x+\frac{3}{15}=\frac{7}{5}\) 

=> \(2x=\frac{7}{5}-\frac{3}{15}=\frac{21}{15}-\frac{3}{15}=\frac{18}{15}\)

=> \(x=\frac{18}{15}:2=\frac{18}{15}\cdot\frac{1}{2}=\frac{9}{15}\cdot\frac{1}{1}=\frac{9}{15}\)

b) \(x-\frac{2}{9}=\frac{8}{3}\)

=> \(x=\frac{8}{3}+\frac{2}{9}\)

=> \(x=\frac{24}{9}+\frac{2}{9}=\frac{26}{9}\)

c) \(\frac{-8}{x}=\frac{-x}{18}\)

=> x(-x) = (-8).18

=> -x² = -144

=> x² = 144(bỏ dấu âm)

=> x = \(\pm\)12

d) \(\frac{2x+3}{6}=\frac{x-2}{5}\)

=> 5(2x + 3) = 6(x - 2)

=> 10x + 15 = 6x - 12

=> 10x + 15 - 6x + 12 = 0

=> 4x + 27 = 0

=> 4x = -27

=> x = -27/4

e) \(\frac{x+1}{22}=\frac{6}{x}\)

=> x(x + 1) = 132

=> x(x + 1) = 11.12

=> x = 11

f) \(\frac{2x-1}{2}=\frac{5}{x}\)

=> x(2x - 1) = 10

=> 2x² - x = 10

=> 2x² - x - 10 = 0

tới đây tự làm đi nhé

g) \(\frac{2x-1}{21}=\frac{3}{2x+1}\)

=> (2x - 1)(2x + 1) = 63

=> 4x² - 1 = 63

=> 4x² = 64

=> x² = 16

=> x = \(\pm\)4

h) Tương tự

a) \(\frac{2x+3}{15}=\frac{7}{5}\Leftrightarrow10x+15=105\Leftrightarrow10x=90\Rightarrow x=9\)

b) \(\frac{x-2}{9}=\frac{8}{3}\Leftrightarrow3x-6=72\Leftrightarrow3x=78\Rightarrow x=26\)

c) \(\frac{-8}{x}=\frac{-x}{18}\Leftrightarrow x^2=144\Leftrightarrow\orbr{\begin{cases}x=12\\x=-12\end{cases}}\)

d) \(\frac{2x+3}{6}=\frac{x-2}{5}\Leftrightarrow10x+15=12x-12\Leftrightarrow2x=27\Rightarrow x=\frac{27}{2}\)

e) \(\frac{x+1}{22}=\frac{6}{x}\Leftrightarrow x^2+x-132=0\Leftrightarrow\left(x-11\right)\left(x+12\right)=0\Leftrightarrow\orbr{\begin{cases}x=11\\x=-12\end{cases}}\)

f) \(\frac{2x-1}{2}=\frac{5}{x}\Leftrightarrow2x^2-x-10=0\Leftrightarrow\left(x-2\right)\left(2x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=2\\x=-\frac{5}{2}\end{cases}}\)

g) \(\frac{2x-1}{21}=\frac{3}{2x+1}\Leftrightarrow4x^2=64\Leftrightarrow x^2=16\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)

h) \(\frac{10x+5}{6}=\frac{5}{x+1}\Leftrightarrow10x^2+15x-25=0\Leftrightarrow5\left(x-1\right)\left(2x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)

Nếu bài dễ thì cố gắng tự làm bạn nhé.

Bài 1:

a) \(2x+3=5\)

\(\Leftrightarrow2x=5-3=2\)

\(\Leftrightarrow x=2:2=1\)

Vậy: x=1

b) \(\left(2x-3\right)^2=9\)\(=3^2=\left(-3\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\left(3+3\right):2=3\\x=\left(-3+3\right):2=0\end{matrix}\right.\)

Vậy: \(x\in\left\{3;0\right\}\)

c) \(3x+4=2x-7\)

\(\Leftrightarrow2x-3x=7+4\)

\(\Leftrightarrow-1x=11\)

\(\Leftrightarrow x=-11\)

Vậy: x=-11

d) \(\left(4x-1\right)^3=27\)\(=3^3\)

\(\Rightarrow4x-1=3\)

\(\Leftrightarrow4x=3+1=4\)

\(\Leftrightarrow x=4:4=1\)

Vậy: x=1

e) \(\left(x-7\right).2=3.\left(2x+4\right)\)

\(\Leftrightarrow2x-14=6x+12\)

\(\Leftrightarrow2x-6x=12+14\)

\(\Leftrightarrow-4x=26\)

\(\Leftrightarrow x=\frac{26}{-4}=\frac{13}{-2}\)

Vậy: \(x=\frac{13}{-2}\)