Bài 3:

a: \(\frac{x}{3}=\frac45\)

=>\(x=3\cdot\frac45=\frac{12}{5}\)

b: (x+2)(x-3)=0

=>\(\left[\begin{array}{l}x+2=0\\ x-3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-2\\ x=3\end{array}\right.\)

c: \(x^2-3x=0\)

=>x(x-3)=0

=>x=0 hoặc x=3

e: \(9^{x}=81\)

=>\(9^{x}=9^2\)

=>x=2

k: 2x=3y=5z

=>\(\frac{2x}{30}=\frac{3y}{30}=\frac{5z}{30}\)

=>\(\frac{x}{15}=\frac{y}{10}=\frac{z}{6}\)

mà x+2y-z=29

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{x}{15}=\frac{y}{10}=\frac{z}{6}=\frac{x+2y-z}{15+2\cdot10-6}=\frac{29}{15+20-6}=\frac{29}{29}=1\)

=>x=15; y=10; z=6

(2-x)^3+(2+x)^3-12x(x+1)=0

=>\(8-12x+6x^2-x^3+8+12x+6x^2+x^3-12x\left(x+1\right)=0\)

=>\(12x^2+16-12x^2-12x=0\)

=>16-12x=0

=>4-3x=0

=>x=4/3

a) (x-2)^x-3(x+1)(x-1)+6x^2=5

<=> \(x^2-4x+4-3(x^2-1)+6x^2-5=0\)

<=>\(x^2-4x+4-3x^2+3+6x^2-5=0\)

<=>\(4x^2-4x+2=0\)

<=> \(4x^2-4x+1+1=0\)

<=>\((2x-1)^2+1=0\)

\(ta\) có \((2x-1)^2 > hoặc = 0\)

             1>0

=> \((2x-1)^2+1=0 (vô lí)\)

=> phuơng trình vô nghiêm S={ rỗng }

a) x³-1-(x²+2x)(x-2)=5

⇔ x³-1-x³+4x=5

⇔ 4x=6

⇔ \(x=\dfrac{3}{2}\)

ghê

a) Ta có bảng sau:

b) Ta có bảng sau:

cảm ơn bn 

a) \(x+1^3=2^5-\left(-1^3\right)\)

\(\Rightarrow x+1=33\)

=> x = 32

b) \(3^7-x=1^4-\left(-3^5\right)\)

\(\Rightarrow2187-x=1+243=244\)

=> x = 1943

a) \(\Leftrightarrow x+1=32+1\)

\(\Leftrightarrow x=32\)

Vậy x = 32

b) \(\Leftrightarrow2187-x=1+243\)

\(\Leftrightarrow2187-x=244\)

\(\Leftrightarrow x=1943\)

Vậy x = 1943

\(\Leftrightarrow x^4+x^3-10x^2+1=x^3-8\)

\(\Leftrightarrow x^4-10x^2+9=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-3\right)\left(x+3\right)=0\)

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

a) \(3\left(x-2\right)+2\left(x-3\right)=5\)

\(\Rightarrow3x-6+2x-6=5\)

\(\Rightarrow5x=17\Rightarrow x=\dfrac{17}{5}\)

b) \(\left(2x-8\right)^2-16=0\)

\(\Rightarrow\left(2x-8-4\right)\left(2x-8+4\right)=0\)

\(\Rightarrow\left(2x-12\right)\left(2x-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x=12\\2x=4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=6\\x=2\end{matrix}\right.\)

c) \(\left(2x-1\right)^2-\left(4x+1\right)\left(x-3\right)=3\)

\(\Rightarrow4x^2-4x+1-4x^2+12x-x+3=3\)

\(\Rightarrow7x=-1\Rightarrow x=-\dfrac{1}{7}\)

a: Ta có: \(3\left(x-2\right)+2\left(x-3\right)=5\)

\(\Leftrightarrow3x-6+2x-6=5\)

\(\Leftrightarrow5x=17\)

hay \(x=\dfrac{17}{5}\)

b: Ta có: \(\left(2x-8\right)^2-16=0\)

\(\Leftrightarrow\left(2x-4\right)\left(2x-12\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)

a: ĐKXĐ: x<>-1/2

\(\dfrac{x-1}{2x+1}=\dfrac{2}{3}\)

=>\(2\left(2x+1\right)=3\left(x-1\right)\)

=>\(4x+2=3x-3\)

=>\(4x-3x=-3-2\)

=>x=-5(nhận)

b: ĐKXĐ: x<>1/2

\(\dfrac{x-2}{2x-1}=\dfrac{-1}{3}\)

=>\(3\left(x-2\right)=-1\left(2x-1\right)\)

=>\(3x-6=-2x+1\)

=>\(3x+2x=1+6\)

=>5x=7

=>x=7/5(nhận)