\(A\left(x\right)=43x-\left(52x^2+34x^2-8x^4\right)-\left(8x^4+16x^3-42x^2+43x\right)+19\)

\(\Leftrightarrow A\left(x\right)=43x-86x^2+8x^4-16x^3+42x^2-43x+19\)

\(\Leftrightarrow A\left(x\right)=-16x^3-44x^2+19\)

Bậc là: 3

8x²+30x+7=0

 8x²+16x+14x+7=0

8x(x+2) +7(x+2)=0

(8x+7)(x+2)=0

=>\(\orbr{\begin{cases}8x+7=0\\x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-\frac{7}{8}\\x=-2\end{cases}}}\)

a)

4x²-8x+4=2(1-x)(x+1)

4x²-8x+4-2+2x²=0

6x²-8x+2=0

2(3x²-4x+1)=0

3x²-3x-x+1=0

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

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

\(\Rightarrow\orbr{\begin{cases}3x-1=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=1\end{cases}}}\)

d,

x²+3x-18=0

=> x²-3x+6x-18=8

=> x(x-3)+6(x-3)=0

=> (x-3)(x+6)=0

=> \(\orbr{\begin{cases}x=3\\x=-6\end{cases}}\)

a) \(8x+56:14=60\)

\(\Rightarrow8x+4=60\)

\(\Rightarrow8x=56\)

\(\Rightarrow x=\dfrac{56}{8}\)

\(\Rightarrow x=7\)

b) Mình làm rồi nhé !

c) \(41-2^{x+1}=9\)

\(\Rightarrow2^{x+1}=41-9\)

\(\Rightarrow2^{x+1}=32\)

\(\Rightarrow2^{x+1}=2^5\)

\(\Rightarrow x+1=5\)

\(\Rightarrow x=4\)

d) \(3^{2x-4}-x^0=8\)

\(\Rightarrow3^{2x-4}-1=8\)

\(\Rightarrow3^{2x-4}=9\)

\(\Rightarrow3^{2x-4}=3^2\)

\(\Rightarrow2x-4=2\)

\(\Rightarrow2x=6\)

\(\Rightarrow x=3\)

g) \(65-4^{x+2}=2014^0\)

\(\Rightarrow65-4^{x+2}=1\)

\(\Rightarrow4^{x+2}=64\)

\(\Rightarrow4^{x+2}=4^3\)

\(\Rightarrow x+2=3\)

\(\Rightarrow x=1\)

i) \(120+2\left(4x-17\right)=214\)

\(\Rightarrow2\left(4x-17\right)=214-120\)

\(\Rightarrow2\left(4x-17\right)=94\)

\(\Rightarrow4x-17=47\)

\(\Rightarrow4x=47+17\)

\(\Rightarrow4x=64\)

\(\Rightarrow x=16\)

a: \(8x+56:14=60\)

=>8x+4=60

=>8x=60-4=56

=>x=56/8=7

b: \(5^{2x-3}-2\cdot5^2=5^2\cdot3\)

=>\(5^{2x-3}=5^2\cdot3+2\cdot5^2=5^3\)

=>2x-3=3

=>2x=6

=>x=3

c: \(41-2^{x+1}=9\)

=>\(2^{x+1}=41-9=32\)

=>x+1=5

=>x=4

d: \(3^{2x-4}-x^0=8\)

=>\(3^{2x-4}-1=8\)

=>\(3^{2x-4}=8+1=9\)

=>2x-4=2

=>2x=6

=>x=3

g: \(65-4^{x+2}=2014^0\)

=>\(65-4^{x+2}=1\)

=>\(4^{x+2}=65-1=64\)

=>x+2=3

=>x=1

i: 120+2(4x-17)=214

=>2(4x-17)=214-120=94

=>4x-17=94/2=47

=>4x=64

=>\(x=\dfrac{64}{4}=16\)

a) Ta có: \(8x^2+30x+7\)

\(=8x^2+28x+2x+7\)

\(=4x\left(2x+7\right)+\left(2x+7\right)\)

\(=\left(2x+7\right)\left(4x+1\right)\)

b) Ta có: \(4x^3-12x^2+9x\)

\(=x\left(4x^2-12x+9\right)\)

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

c) Ta có: \(\left(2x+1\right)^2-\left(x-1\right)^2\)

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

\(=\left(x+2\right)\cdot3x\)

d) Ta có: \(ab+c^2-ac-bc\)

\(=\left(ab-bc\right)+\left(c^2-ac\right)\)

\(=b\left(a-c\right)+c\left(c-a\right)\)

\(=b\left(a-c\right)-c\left(a-c\right)\)

\(=\left(a-c\right)\left(b-c\right)\)

e) Ta có: \(4x^2-y^2+1-4x\)

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

\(=\left(2x-1\right)^2-y^2\)

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

f) Ta có: \(6x^2-7x-20\)

\(=6x^2-15x+8x-20\)

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

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

\(4x^3-12x^2+9x=x\left(4x^2-12x+9\right)=x\left(2x-3\right)^2\), \(\left(2x+1\right)^2-\left(x-1\right)^2=\left(2x+1-x+1\right)\left(2x+1+x-1\right)=\left(x+2\right)3x\)

\(ab+c^2-ac-bc=ab-ac-bc+c^2=a\left(b-c\right)-c\left(b-c\right)=\left(b-c\right)\left(a-c\right)\)

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

\(6x^2-7x-20=6x^2-15x+8x-20=3x\left(2x-5\right)+4\left(2x-5\right)=\left(2x-5\right)\left(3x+4\right)\)

\(8x^2+30x+7=8x^2+2x+28x+7=2x\left(4x+1\right)+7\left(4x+1\right)=\left(4x+1\right)\left(2x+7\right)\)

\(=8x^2+2x+28x+7=2x\left(4x+1\right)+7\left(4x+1\right)=\left(2x+7\right)\left(4x+1\right)\)