\(\left\{{}\begin{matrix}4x=3y\\4x+4y=4\cdot14=56\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x=3y\\3y+4y=56\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x=3y\\y=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=6\\y=8\end{matrix}\right.\)

\(4x=3y=>y=\dfrac{4x}{3}\) thế y vào x+y=14

\(=>x+\dfrac{4x}{3}=14< =>\dfrac{7x}{3}=14=>7x=42< =>x=6\)

\(=>y=\dfrac{4.6}{3}=8\)

4x=3y tương đương x/3=y/4

Áp dụng.............

x/3=y/4=x+y/3+4=14/7=2

\(\Leftrightarrow\left[{}\begin{matrix}x-3=1\\x-3=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

a) \(A\left(x\right)=3x^3-4x^4-2x^3+4x^4-5x+3\)

\(\Rightarrow A\left(x\right)=-4x^4+4x^4+3x^3-2x^3-5x+3\)

\(\Rightarrow A\left(x\right)=x^3-5x+3\)

\(B\left(x\right)=5x^3-4x^2-5x^3-4x^2-5x-3\)

\(\Rightarrow B\left(x\right)=5x^3-5x^3-4x^2-4x^2-5x-3\)

\(\Rightarrow B\left(x\right)=-8x^2-5x-3\)

b) \(A\left(x\right)+B\left(x\right)=x^3-5x+3+\left(-8x^2-5x-3\right)\)

\(\Rightarrow A\left(x\right)+B\left(x\right)=x^3-5x+3-8x^2-5x-3\)

\(\Rightarrow A\left(x\right)+B\left(x\right)=x^3-8x^2-5x-5x+3-3\)

\(\Rightarrow A\left(x\right)+B\left(x\right)=x^3-8x^2-10x\)

\(A\left(x\right)-B\left(x\right)=x^3-5x+3-\left(-8x^2-5x-3\right)\)

\(\Rightarrow A\left(x\right)-B\left(x\right)=x^3-5x+3+8x^2+5x+3\)

\(\Rightarrow A\left(x\right)-B\left(x\right)=x^3+8x^2-5x+5x+3+3\)

\(\Rightarrow A\left(x\right)-B\left(x\right)=x^3+8x^2+6\)

A(x) = 3x^3 - 4x^4 - 2x^3 + 4x^4 - 5x + 3

= (4x^4 - 4x^4) + (3x^3 - 2x^3) - 5x + 3

= x^3 - 5x + 3

B(x) = 5x^3 - 4x^2 - 5x^3 - 4x^2 - 5x - 3

= (5x^3 - 5x^3) - (4x^2 + 4x^2) - 5x - 3

= -8x^2 - 5x - 3

b.

\(A\left(x\right)+B\left(x\right)=x^3-5x+3-8x^2-5x-3\) 

\(=x^3-8x^2+\left(-5x-5x\right)+\left(3-3\right)\) 

\(=x^3-8x^2-10x\)

\(A\left(x\right)-B\left(x\right)=\left(x^3-5x+3\right)-\left(-8x^2-5x-3\right)\) 

\(=x^3-5x+3+8x^2+5x+3\) 

\(=x^3-8x^2+\left(-5x+5x\right)+\left(3+3\right)\)

\(=x^3-8x^2+6\) 

Vậy \(A\left(x\right)+B\left(x\right)=x^3-8x^2-10x\) 

      \(A\left(x\right)-B\left(x\right)=x^3-8x^2+6\)

\(A\left(x\right)=x^3-5x+3\)

Vừa thu gọn vừa sắp xếp nhé

Hok tốt

a)A(x) = 3x^3 - 4x^4 - 2x^3 + 4x^4 - 5x + 3 

=x^3-5x+3

bậc:3

hệ số tự do:3

hệ số cao nhất :3

B(x) = 5x^3 - 4x^2 - 5x^3 - 4x^2 - 5x - 3

=-8x^2-5x+3

bậc:2

hệ số tự do:3

hệ số cao nhất:3

b)A(x)+B(x)=x^3-8^2+10x+6

câu b mik ko đặt tính theo hàng dọc đc thông cảm nha

a, \(4\left(18-5x\right)-12\left(3x-7\right)=15\left(2x-16\right)-6\left(x+14\right)\)

\(\Rightarrow72-20x-36x+84=30x-240-6x-84\)

\(\Rightarrow-20x-36x-30x+6x=-240-84-72-84\)

\(\Rightarrow-80x=-480\Rightarrow x=6\)

b, \(5\left(3x+5\right)-4\left(2x-3\right)=5x+3\left(2x+12\right)+1\)

\(\Rightarrow15x+25-8x+12=5x+6x+36+1\)

\(\Rightarrow15x-8x-5x-6x=36+1-25-12\)

\(\Rightarrow-4x=0\Rightarrow x=0\)

c, \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Rightarrow10x-16-12x+15=12x-16+11\)

\(\Rightarrow10x-12x-12x=-16+11+16-15\)

\(\Rightarrow-14x=-4\Rightarrow x=\dfrac{2}{7}\)

d, \(5x-3\left\{4x-2\left[4x-3\left(5x-2\right)\right]\right\}=182\)

\(\Rightarrow5x-3\left[4x-2\left(4x-15x+6\right)\right]=182\)

\(\Rightarrow5x-3\left(4x-8x+30x-12\right)=182\)

\(\Rightarrow5x-12x+24x-90x+36=182\)

\(\Rightarrow-73x=182-36\)

\(\Rightarrow-73x=146\Rightarrow x=-2\)

Chúc bạn học tốt!!!

a) \(\left(5x-2\right)\left(5x+2\right)-\left(5x+3\right)\left(5x-4\right)=0\)

\(\Leftrightarrow5x+8=8\)

\(\Leftrightarrow5x=8-8\)

\(\Leftrightarrow x=5.0\)

\(\Leftrightarrow x=0\)

b) 

câu b sao bạn

\(\left(4x-3\right)\left(4x+2\right)+\left(4x+5\right)\left(1-4x\right)=2.5^2\)

\(16x^2+8x-12x-6+4x-16x^2+5-20x=50\)

\(-20x-1=50\)

\(-20x=51\)

\(x=\frac{-51}{20}\)

Vậy \(x=\frac{-51}{20}\)

a: \(=2x^3:\dfrac{-3}{2}x+4x:\dfrac{3}{2}x-5:\dfrac{3}{2}\)

=-4/3x^2+8/3-10/3

=-4/3x^2-2/3

d: \(\dfrac{3x^3-5x+2}{x-3}=\dfrac{3x^3-9x^2+9x^2-27x+22x-66+68}{x-3}\)

\(=3x^2+9x+22+\dfrac{68}{x-3}\)