Question

tìm x
a) x^2 - 100 - (x+10) = 0
b) x^2.(x^2 + 2) - x^2 - 2 =0
c) (3x - 1)^2 - (9x^2 -1 )=0
làm hết hộ nha thank

Answer 1

a) \(x^2-100-\left(x+10\right)=0\)

\(x^2-x-90=0\)

\(\left(x-10\right)\left(x+9\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=10\\x=-9\end{matrix}\right.\)

b) \(x^2.\left(x^2+2\right)-x^2-2=0\)

\(x^2\left(x^2+2\right)-\left(x^2+2\right)=0\)

\(\left(x^2+2\right)\left(x^2-1\right)=0\)

\(\left(x^2+2\right)\left(x-1\right)\left(x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x^2=-2\\x=1\\x=-1\end{matrix}\right.\)

Loại trường hợp \(x^2=-2\) vì x² \(\ge\) 0 với mọi x

c) \(\left(3x-1\right)^2-\left(9x^2-1\right)=0\)

\(\left(3x-1\right)^2-\left(3x-1\right)\left(3x+1\right)=0\)

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

\(-2.\left(3x-1\right)=0\)

\(3x-1=0\Rightarrow x=\dfrac{1}{3}\)

Answer 2

a,x²-100-(x+10)=0 b,x²(x²+2)-x²-2=0

(x-10)(x+10)-(x+10)=0 x²(x²+2)-(x²+2)=0

(x+10)(x-10-1)=0 (x²-1)(x²+2)=0

(x+10)(x-11)=0 (x-1)(x+1)(x²+2)=0

<=>1)x+10=0 <=>1)x=-10 <=>1)x-1=0 <=>1)x=1

2)x-11=0 2)x=11 2)x+1=0 2)x= -1

Vậy... 3)x²+2=0 3)x²= -2(vô lí)

Vậy....

c,(3x-1)²-(9x²-1)=0

(3x-1)²-(3x-1)(3x+1)=0

(3x-1)(3x-1-3x-1)=0

(3x-1)(-2)=0

<=>3x-1=0

<=>x=\(\dfrac{1}{3}\)

Vậy ....