a: \(\Leftrightarrow\left(4x+12\right)\left(3x-2\right)-\left(3x+3\right)\left(4x-1\right)=-27\)

\(\Leftrightarrow12x^2-8x+36x-24-\left(12x^2-3x+12x-3\right)=-27\)

\(\Leftrightarrow12x^2+28x-24-12x^2-9x+3=-27\)

\(\Leftrightarrow19x-21=-27\)

=>19x=-6

hay x=-6/19

b: \(\left(x+1\right)\left(3x^2-x+1\right)+x^2\left(4-3x\right)=\dfrac{5}{2}\)

\(\Leftrightarrow3x^3-x^2+x+3x^2-x+1+4x^2-3x^3=\dfrac{5}{2}\)

\(\Leftrightarrow6x^2+1=\dfrac{5}{2}\)

\(\Leftrightarrow6x^2=\dfrac{3}{2}\)

\(\Leftrightarrow x^2=\dfrac{3}{12}=\dfrac{1}{4}\)

=>x=1/2 hoặc x=-1/2

c: \(\Leftrightarrow2\left(x^2-4\right)-4\left(x^2-x-2\right)+\left(5x+8\right)\left(x+2\right)=0\)

\(\Leftrightarrow2x^2-8-4x^2+4x+8+5x^2+10x+8x+16=0\)

\(\Leftrightarrow3x^2+22x+16=0\)

\(\text{Δ}=22^2-4\cdot3\cdot16=292>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-22-2\sqrt{73}}{6}=\dfrac{-11-\sqrt{73}}{3}\\x_2=\dfrac{-11+\sqrt{73}}{3}\end{matrix}\right.\)

d: \(\Leftrightarrow20x^2-16x-1=10x^2-2x+5x-1\)

\(\Leftrightarrow10x^2-19x=0\)

=>x(10x-19)=0

=>x=0 hoặc x=19/10

a, 15/x - 1/3 = 28/57

15/x = 28/57 + 1/3

15/x = 28/57 + 19/57

15/x = 47/57

x . 47 = 15 . 57

x = 855/47

b, x/2 - 2/5 = 1/10

x/2 = 1/10 + 2/5 

x/2 = 1/10 + 4/10

x/2 = 5/10 = 1/2

x/2 = 1/2

=> x=1

\(\left(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}\right):\dfrac{4x}{10x-5}\)

\(=\left(\dfrac{2x+1}{-\left(2x+1\right)}-\dfrac{2x-1}{-\left(2x-1\right)}\right):\dfrac{4x}{10x-5}\)

\(=\dfrac{2x}{5x-5}\)

Grandpa was slept, visted,arrived

Grandpa was sleeping when the visitors arrived.

2 câu dễ làm trước, 2 câu còn lại tối đi học về mới làm được..(giờ bận rồi)

a) ĐẶt \(x^2+3x+1=a\)

\(A=a\left(a-4\right)-5=a^2-4a-5=\left(a-5\right)\left(a+1\right)\)

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

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

c)\(C=\left[\left(x+1\right)\left(x+7\right)\right]\left[\left(x+3\right)\left(x+5\right)\right]+15\)

\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)

Đặt ẩn phụ: \(t=x^2+8x+7\) rồi làm tiếp đi..

Để anh làm nốt vậy.

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

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

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

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

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

___

\(D=x^2-2xy+y^2-7x+7y+12\)

\(D=\left(x-y\right)^2-7\left(x-y\right)+12\)

\(D=\left(x-y\right)^2-3\left(x-y\right)-4\left(x-y\right)+12\)

\(D=\left(x-y\right)\left(x-y-3\right)-4\left(x-y-3\right)\)

\(D=\left(x-y-3\right)\left(x-y-4\right)\)

Bạn gõ ct lại dưới dạng latex (kí hiệu này gọi là latex Σ ) nhé