a) \(=x^2-49-x^2\) \(=-49\)

b) \(=25x^2-1-25x^2-1\) \(=-2\)

c) \(=16x^2-1-16x^2+8x-1\) \(=8x-2\)

d) \(=9x^2-30x+25-9x^2+25\) \(=50-30x\)

ai giúp mình với

plz

tham khảo

f(x) = x5 + 3x2 − 5x3 − x7 + x3 + 2x2 + x5 − 4x2 + x7

= (x5 + x5) + (3x2 + 2x2 – 4x2) + (-5x3 + x3) + (-x7 + x7)

= 2x5 + x2 – 4x3.

= 2x5 - 4x3 + x2

Đa thức có bậc là 5

g(x) = x4 + 4x3 – 5x8 – x7 + x3 + x2 – 2x7 + x4 – 4x2 – x8

= (x4 + x4) + (4x3 + x3) – (5x8 + x8) – (x7 + 2x7) + (x2 – 4x2)

= 2x4 + 5x3 – 6x8 – 3x7 – 3x2

= -6x8 - 3x7 + 2x4 + 5x3 - 3x2.

Đa thức có bậc là 8.

\(5+4x-x+2=\left(5x+4\right)\left(7+5x\right)\)

\(\Leftrightarrow5+4x-x+2=35+28x+25x+20x^2\)

\(\Leftrightarrow x^2+50x+28=0\)

Ta có \(\Delta=50^2-4.1.28=2388,\sqrt{\Delta}=2\sqrt{597}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-50+2\sqrt{597}}{2}=-25+\sqrt{597}\\x=\frac{-50-2\sqrt{597}}{2}=-25-\sqrt{597}\end{cases}}\)

\(5+4x-x+2=\left(5+4x\right)\left(7+5x\right)\)

\(7+3x=\left(5+4x\right)\left(7+5x\right)\)

\(7+3x=35+28x+25x+20x^2\)

\(7+3x-35-28x-25x-20x^2=0\)

\(-28-50x-20x^2=0\)

\(-28-50x-20x^2=0\)

\(x=-\frac{25+\sqrt{65}}{20};-\frac{25-\sqrt{65}}{20}\)

\(5+4x-x+2=\left(5+4x\right)\left(7+5x\right)\)

\(7+3x=35+25x+28x+20x^2\)

\(7+3x=35+53x+20x^2\)

\(7+3x-35-53x-20x^2=0\)

\(-28-50x-20x^2=0\)

\(x=-\frac{25+\sqrt{65}}{20};-\frac{25-\sqrt{65}}{20}\)

     5+4x-x+2=(5+4x).(7+5x)

(5+2)+(4x-x)=(5+4x).(7+5x)

            7+3x=(5+4x).(7+5x)

\(\Rightarrow\orbr{\begin{cases}5+4x=7+3x\\7+5x=7+3x\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}\)

Vậy x=2 hoặc x=0

A = - 3\(x\).(\(x-5\)) + 3(\(x^2\) - 4\(x\)) - 3\(x\) - 10

A = - 3\(x^2\) + 15\(x\) + 3\(x^2\) - 12\(x\) - 3\(x\) - 10

A = (- 3\(x^2\) + 3\(x^2\)) + (15\(x\) - 12\(x\) - 3\(x\)) - 10

A = 0 + (3\(x-3x\)) - 10

A = 0  - 10

A = - 10 

a: Ta có: \(\sqrt{4x^2+4x+3}=8\)

\(\Leftrightarrow4x^2+4x+1+2-64=0\)

\(\Leftrightarrow4x^2+4x-61=0\)

\(\Delta=4^2-4\cdot4\cdot\left(-61\right)=992\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-4-4\sqrt{62}}{8}=\dfrac{-1-\sqrt{62}}{2}\\x_2=\dfrac{-4+4\sqrt{62}}{8}=\dfrac{-1+\sqrt{62}}{2}\end{matrix}\right.\)

\(a,5x^2-3x\left(x-2\right)\)

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

\(=2x^2+6x\)
\(b,3x\left(x-5\right)-5x\left(x+7\right)\)
\(=3x^2-15x-5x^2-35x\)

\(=-2x^2-50x\)

c, Đề ko rõ Yang Yang

\(d,7x\left(x-5\right)+3\left(x-2\right)\)

\(=7x^2-35x+3x-6\)

\(=7x^2-32x-6\)

\(e,5-4x\left(x-2\right)+4x^2\)

\(=5-4x^2+8x+4x^2\)

\(=5+8x\)

\(f,4x\left(2x-3\right)-5x\left(x-2\right)\)

\(=8x^2-12x-5x^2+10x\)

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

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!!!