\(40+2\left(12-x\right)=60\)

\(2\left(12-x\right)=20\)

\(12-x=10\)

\(x=2\)

\(40+2.\left(12-x\right)=60\)

\(40+24-2x=60\)

\(64-2x=60\)

\(2x=4\)

\(x=2\)

c: \(\Leftrightarrow4^x\cdot15=60\)

hay x=1

Đề bài yêu cầu gì?

tính nhanh nhe 

1) 2x – (3 – 5x) = 4( x +3)

<=>2x-3+5x=4x+12

<=>2x-3+5x-4x-12=0

<=>3x-15=0

<=>x=5

2) 5(2x-3) - 4(5x-7) =19 - 2(x+11)

<=>10x-15-20x+28=19-2x-22

<=>10x-15-20x+28-19+2x+22=0

<=>-8x+16=0

<=>x=2

\(\left(5x+2\right)\left(x-1\right)-3\left(x+3\right)^2-2\left(x-6\right)\left(x+6\right)\)

\(=5x^2-5x+2x-2-3\left(x^2+6x+9\right)-2\left(x^2-6^2\right)\)

\(=5x^2-3x-2-3x^2-18x-27-2x^2+72\)

\(=-21x+43\)

kết quả mik ra là -3x + 7 theo các bạn đúng ko

Ta có: \(\left(x^2-1\right)\left(x^2+5x+6\right)=60\)

=>\(x^4+5x^3+6x^2-x^2-5x-6-60=0\)

=>\(x^4+5x^3+5x^2-5x-66=0\)

=>\(x^4-2x^3+7x^3-14x^2+19x^2-38x+33x-66=0\)

=>(x-2)(\(x^3+7x^2+19x+33)\) =0

TH1: x-2=0

=>x=2

TH2: \(x^3+7x^2+19x+33=0\)

=>x≃-4,38

\(\lim\limits_{x\rightarrow-\infty}\dfrac{x-5x^2+1}{x^2-1}=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{1}{x}-5+\dfrac{1}{x^2}}{1-\dfrac{1}{x^2}}=\dfrac{-5}{1}=-5\)

\(\lim\limits_{x\rightarrow+\infty}\dfrac{5x^3\left(2-x^2\right)^3\left(4x^2+1\right)^2}{4x^{13}+x^2-6}=\lim\limits_{x\rightarrow+\infty}\dfrac{5\left(\dfrac{2}{x^2}-1\right)^3\left(4+\dfrac{1}{x^2}\right)^2}{4+\dfrac{1}{x^{11}}-\dfrac{6}{x^{13}}}=\dfrac{5.\left(-1\right)^3.4^2}{4}=-20\)

\(\lim\limits_{x\rightarrow+\infty}\dfrac{4x-\sqrt{9x^2+x}}{3-x}=\lim\limits_{x\rightarrow+\infty}\dfrac{4-\sqrt{9+\dfrac{1}{x}}}{\dfrac{3}{x}-1}=\dfrac{4-3}{-1}=-1\)