\(Câu8\)

\(a,A=\dfrac{1}{2}x^3\times\dfrac{8}{5}x^2=\left(\dfrac{1}{2}\times\dfrac{8}{5}\right)x^{3+2}=\dfrac{4}{5}x^5\)

b, \(P\left(0\right)=0^2-5.0+6=6\\ P\left(2\right)=2^2-5.2+6=0\)

Câu 9

\(a,A\left(x\right)+B\left(x\right)=5x^3+x^2-3x+5+5x^3+x^2+2x-3\\ =\left(5x^3+5x^3\right)+\left(x^2+x^2\right)+\left(-3x+2x\right)+\left(5-3\right)\\ =10x^3+2x^2-x+2\)

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

\(H\left(x\right)=0\\ \Rightarrow-5x+8=0\\ \Rightarrow x=\dfrac{8}{5}\)

vậy nghiệm của đa thức là \(x=\dfrac{8}{5}\)

a; Bậc của A(x) là 3

Bậc của B(x) là 3

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

\(A\left(x\right)-B\left(x\right)=-x^3-5x^2+3x+7\)

c: C(x)=0

=>5x-15=0

hay x=3

`a)`

`@` Bậc của `A(x)` là: `3`

`@` Bậc của `B(x)` là: `3`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`b)`

`@A(x)+B(x)=2x^3-3x^2+2x+1+3x^3+2x^2-x-6`

                   `=5x^3-x2+x-5`

`@A(x)-B(x)=2x^3-3x^2+2x+1-3x^3-2x^2+x+6`

                  `=-x^3-5x^2+3x+7`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`c)` Cho `C(x)=0`

`=>5x-15=0`

`=>5x=15`

`=>x=3`

Vậy nghiệm của `C(x)` là `x=3`

Tham khảo:

a. Bậc của A(x) là 3

Bậc của B(x) là 3

b. $A\left(x\right)+B\left(x\right)=5{}^{}-{}^{}+x-5$

$A\left(x\right)-B\left(x\right)=-{}^{}-5{}^{}+3x+7$

c. C(x)= 0

=>5x - 15= 0

hay x= 3

a: 3x-5>15-x

=>4x>20

hay x>5

b: \(3\left(x-2\right)\left(x+2\right)< 3x^2+x\)

=>3x²+x>3x²-12

=>x>-12

Bài 2:

a: A(x)=0

=>-4x+7=0

=>4x=7

=>x=7/4

b: B(x)=0

=>x(x+2)=0

=>x=0 hoặc x=-2

c: C(x)=0

=>1/2-căn x=0

=>căn x=1/2

=>x=1/4

d: D(x)=0

=>2x^2-5=0

=>x^2=5/2

=>\(x=\pm\dfrac{\sqrt{10}}{2}\)

\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)

\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)

Bài 4:

a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)

\(\Leftrightarrow6x-9-2x+4=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

hay \(x=\dfrac{13}{3}\)

c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

hay x=1

a/ \(3\left(2x-3\right)+2\left(2-x\right)=-3\)

\(\Leftrightarrow6x-9+4-2x=-3\)

\(\Leftrightarrow4x=2\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy: \(x=\dfrac{1}{2}\)

===========

b/ \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

\(\Leftrightarrow x=\dfrac{13}{3}\)

Vậy: \(x=\dfrac{13}{3}\)

==========

c/  \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

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

\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=8\)

\(\Leftrightarrow-2x=-2\)

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

e/ \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow-14x=-4\)

\(\Leftrightarrow x=\dfrac{2}{7}\)

Vậy: \(x=\dfrac{2}{7}\)

==========

f/ \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)

\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)

\(\Leftrightarrow-x^3=8\)

\(\Leftrightarrow x=-2\)

Vậy: \(x=-2\)

bạn cho các đa thức A(x) , B(x),C(x) =0 rồi giải nha !

\(A(x)=(2x-4)(x+1)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-4=0\\x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)

Vậy đa thức \(A(x)\)có hai nghiệm đó là 2 và -1

\(B(x)=(-5x+2)(x-7)=0\)

\(\Leftrightarrow\orbr{\begin{cases}-5x+2=0\\x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}-5x=-2\\x=7\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{5}\\x=7\end{cases}}\)

Vậy đa thức \(B(x)\)có hai nghiệm đó là 2/5 và 7

\(C(x)=(4x-3)(2x+3)=0\)

\(\Leftrightarrow\orbr{\begin{cases}4x-3=0\\2x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=-\frac{3}{2}\end{cases}}\)

Vậy đa thức \(C(x)\)có hai nghiệm đó là 3/4 và -3/2

1) (2x - 4)(x + 1) = 0

2x - 4 = 0 hoặc x + 1 = 0

2x = 0 + 4         x = 0 - 1

2x = 4               x = -1

x = 4/2

x = 2

=> x = 2 hoặc x = -1

2) (-5x + 2)(x - 7) = 0

-5x + 2 = 0 hoặc x - 7 = 0

-5x = 0 - 2           x = 0 + 7

-5x = -2               x = 7

x = 2/5                 x = 7

=> x = 2/5 hoặc x = 7

3) (4x - 3)(2x + 3) = 0

4x - 3 = 0 hoặc 2x + 3 = 0

4x = 0 + 3         2x = 0 - 3

4x = 3               2x = -3

x = 3/4              x = -3/2

=> x = 3/4 hoặc x = -3/2

a. \(5x\left(2x-7\right)+2x\left(8-5x\right)=5\)

\(\Rightarrow10x^2-35x+16x-10x^2=5\)

\(\Rightarrow-19x=5\)

\(\Rightarrow x=-\frac{5}{19}\)

b. \(x\left(x-\frac{1}{3}\right)-\frac{1}{2}x\left(2x-3\right)=\frac{1}{4}\)

\(\Rightarrow x^2-\frac{1}{3}x-x^2+\frac{3}{2}x=\frac{1}{4}\)

\(\Rightarrow\frac{7}{6}x=\frac{1}{4}\)

\(\Rightarrow x=\frac{3}{14}\)

c. \(5\left(x^2-3x+1\right)+x\left(1-5x\right)=x-2\)

\(\Rightarrow5x^2-15x+5+x-5x^2=x-2\)

\(\Rightarrow-14x+5=x-2\)

\(\Rightarrow-14x-x=-2-5\)

\(\Rightarrow-15x=-7\)

\(\Rightarrow x=\frac{7}{15}\)

a, \(5x\left(2x-7\right)+2x\left(8-5x\right)=5\)

\(\Leftrightarrow10x^2-35x+16x-10x^2=5\)

\(\Leftrightarrow-19x=5\Leftrightarrow x=-\frac{5}{19}\)

b, \(x\left(x-\frac{1}{3}\right)-\frac{1}{2}x\left(2x-3\right)=\frac{1}{4}\)

\(\Leftrightarrow x^2-\frac{1}{3}x-x^2+\frac{3}{2}x=\frac{1}{4}\)

\(\Leftrightarrow\frac{7}{6}x=\frac{1}{4}\Leftrightarrow x=\frac{3}{14}\)

c, \(5\left(x^2-3x+1\right)+x\left(1-5x\right)=x-2\)

\(\Leftrightarrow5x^2-15x+5+x-5x^2=x-2\)

\(\Leftrightarrow-15x+7=0\Leftrightarrow x=\frac{7}{15}\)