Bạn ơi tìm x nhưng x thuộc j đấy hay tìm x ko thôi

a) (3x + 5)² = 289

Vì \(\sqrt{289}\)= 17

=> 3x + 5 = 17

3x = 17 - 5

3x = 12

x = 12 : 3

x = 4

b) x + (x²)³ = x⁵

x + x⁶ = x⁵

x = x⁵ - x⁶

x = x^-1 = \(\frac{1}{x}\).

a,x².x³=2⁵

=>x⁵=2⁵

=>x=2

b,x+18=5.4^2

=>x=5.16-18

=>x=62

c,x.(x^2)^3=x^5

=>x.x⁵=x⁵

=>x=0,1

d,2^x.7=224

2^x=32

=>2^x=2⁵

=>x=5

e,(3x+5)²=289

=>(3x+5)²=17²

=>3x+5=17

=>3x=12

=>x=4

g,3^2x+1.11=2673

=>3^2x=243

=>3^2x=3⁵

=>x=\(\frac{5}{2}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^3+2y^2-4y+3=0\\2x^2+2x^2y^2-4y=0\left(1\right)\end{matrix}\right.\Rightarrow}x^3+2y^2-4y-2x^2-2x^2y^2+4y=0\Rightarrow x^3+1-2x^2y^2+2y^2-2x^2+2=0\Rightarrow\left(x+1\right)\left(x^2-x+1\right)-2y^2\left(x-1\right)\left(x+1\right)-2\left(x-1\right)\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(x^2-x+1-2xy^2+2y^2-2x+2\right)=0\Rightarrow x=-1\)Thay x=-1 vào (1) ta được y²-2y+1=0⇒ (y-1)²=0⇒y-1=0⇒y=1

Do đó Q=x²+y²=(-1)²+1²=2

\(B=\frac{5}{2.1}+\frac{1}{11}.\left(4+\frac{3}{2}\right)+\frac{1}{15}.\left(\frac{1}{2}+\frac{1}{4}\right)\)

\(B=\frac{5}{2.1}+\frac{1}{11}.5\frac{1}{2}+\frac{1}{15}.\frac{3}{4}\)

tự giải tiếp nhé

`x^2-2x-sqrt3+1=0`
Vì `Delta=1+sqrt3-1>0`
`=>` pt có 2 nghiệm pb
ÁP dụng vi-ét:
`x_1+x_2=2,x_1.x_2=1-sqrt3`
`M=x_1^2x_2^2-2x_1.x_2-x_1-x_2`
`=(x_1.x_2)^2-2(x_1.x_2)-(x_1+x_2)`
`=(sqrt3-1)^2-2(1-sqrt3)-2`
`=4-2sqrt3-2+2sqrt3-2`
`=0`