Cho đa thức:
\(A=3x-2-\left|2x+1\right|\)
Rút gọn A
Rút gọn biểu thức sau:
A=\(\left(2x+y\right)^2-\left(y-2x\right)^2\)
B=\(\left(3x+2\right)^2+2\cdot\left(2+3x\right)\cdot\left(1-2y\right)+\left(2y-1\right)^2\)
a: Ta có: \(A=\left(2x+y\right)^2-\left(2x-y\right)^2\)
\(=\left(2x+y-2x+y\right)\left(2x+y+2x-y\right)\)
\(=4x\cdot2y=8xy\)
b: Ta có: \(B=\left(3x+2\right)^2+2\left(3x+2\right)\left(1-2y\right)+\left(2y-1\right)^2\)
\(=\left(3x+2+1-2y\right)^2\)
\(=\left(3x-2y+3\right)^2\)
Rút gọn đa thức : \(\left(2x^2-3x+7\right)-\left(3x^2-5x+4\right)-2x+x^2\)
Chị @Hoàng Lê Bảo Ngọc chỉ cho đáp án nha
Giuos nha
(2x2 - 3x + 7) - (3x2 - 5x + 4) - 2x + x2
= 2x2 - 3x + 7 - 3x2 + 5x - 4 - 2x + x2
= 3
Rút gọn đa thức
\(\left(2x+3\right)^2+\left(2x-3\right)^2+2\left(1-2x\right)\left(2x-1\right)\)
\(\left(2x+3\right)^2+\left(2x-3\right)^2+2\left(1-2x\right)\left(2x-1\right)\)
\(=4x^2+12x+9+4x^2-12x+9-2\left(2x-1\right)^2\)
\(=8x^2+18-2\left(4x^2-4x+1\right)\)
\(=8x^2+18-8x^2+8x-2=8x+16\)
Rút gọn các biểu thức sau :
A = \(2x^2\left(-3x^3+2x^2+x-1\right)+2x\left(x^2-3x+1\right)\)
B = \(2x:\dfrac{1}{2}x+x^2\)
C = \(\left[1:\left(1+x\right)+2x:\left(1-x^2\right)\right]:\left(\dfrac{1}{x}-1\right)\)
D = \(\dfrac{x^2-y^2}{x+y}.\dfrac{\left(x+y\right)^2}{x}+\dfrac{y^2}{x+y}.\dfrac{\left(x+y\right)^2}{x}\)
E = \(\dfrac{\left|x-3\right|}{x^2-9}.\left(x^2+6x+9\right)\)
F = \(\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{10\sqrt{x}}{x-25}-\dfrac{5}{\sqrt{x}+5}\)
Rút gọn các biểu thức sau :
A = \(2x^2\left(-3x^3+2x^2+x-1\right)+2x\left(x^2-3x+1\right)\)
B = \(2x:\dfrac{1}{2}x+x^2\)
C = \(\left[1:\left(1+x\right)+2x:\left(1-x^2\right)\right]:\left(\dfrac{1}{x}-1\right)\)
D = \(\dfrac{x^2-y^2}{x+y}.\dfrac{\left(x+y\right)^2}{x}+\dfrac{y^2}{x+y}.\dfrac{\left(x+y\right)^2}{x}\)
E = \(\dfrac{\left|x-3\right|}{x^2-9}.\left(x^2+6x+9\right)\)
F = \(\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{10\sqrt{x}}{x-25}-\dfrac{5}{\sqrt{x}+5}\)
1.Phân tích đa thức thành nhân tử
a.\(2x^3+3x^2-2x\) b.\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)
2.Cho A=\(\dfrac{2x+1}{\left(x-4\right)\left(x-3\right)}-\dfrac{x+3}{x-4}+\dfrac{2x-1}{x-3}\)
a.Rút gọn biểu thức A
b.tính giá trị của A biết \(x^2+20=9x\)
3.Tìm đa thức thương và đa thức dư trong phép chia:\(\left(2x^2-7x^2:13x:2\right):\left(2x-1\right)\)
Bài 1:
a. $2x^3+3x^2-2x=2x(x^2+3x-2)=2x[(x^2-2x)+(x-2)]$
$=2x[x(x-2)+(x-2)]=2x(x-2)(x+1)$
b.
$(x+1)(x+2)(x+3)(x+4)-24$
$=[(x+1)(x+4)][(x+2)(x+3)]-24$
$=(x^2+5x+4)(x^2+5x+6)-24$
$=a(a+2)-24$ (đặt $x^2+5x+4=a$)
$=a^2+2a-24=(a^2-4a)+(6a-24)$
$=a(a-4)+6(a-4)=(a-4)(a+6)=(x^2+5x)(x^2+5x+10)$
$=x(x+5)(x^2+5x+10)$
Bài 2:
a. ĐKXĐ: $x\neq 3; 4$
\(A=\frac{2x+1-(x+3)(x-3)+(2x-1)(x-4)}{(x-3)(x-4)}\\ =\frac{2x+1-(x^2-9)+(2x^2-9x+4)}{(x-3)(x-4)}\\ =\frac{x^2-7x+14}{(x-3)(x-4)}\)
b. $x^2+20=9x$
$\Leftrightarrow x^2-9x+20=0$
$\Leftrightarrow (x-4)(x-5)=0$
$\Rightarrow x=5$ (do $x\neq 4$)
Khi đó: $A=\frac{5^2-7.5+14}{(5-4)(5-3)}=2$
Bài 3:
$(2x^2-7x^2:13x:2):(2x-1)=(2x^2-\frac{7}{26}x):(2x-1)$
$=[x(2x-1)+\frac{19}{52}(2x-1)+\frac{19}{52}]:(2x-1)$
$=[(2x-1)(x+\frac{19}{52})+\frac{19}{52}]: (2x-1)$
$\Rightarrow$ thương là $x+\frac{19}{52}$ và thương là $\frac{19}{52}$
Rút gọn biểu thức:
a/ \(\left(x+2\right)\left(x-2\right)-\left(x-3\right)\left(x+1\right)\)
b/ \(\left(2x+1\right)^2+\left(3x-1\right)^2+2\left(2x+1\right)\left(3x-1\right)\)
(x+2)(x-2)-(x-3)(x+1)
=x^2-2x+2x-4-x^2-x-3x-3
=-4x-7
Rút gọn các biểu thức sau:
a)
\(2x\left(2x-1\right)^2-3x\left(x+3\right)\left(x-3\right)-4x\left(x+1\right)^2\)
\(2x\left(2x-1\right)^2-3x\left(x+3\right)\left(x-3\right)-4x\left(x+1\right)^2\)
\(=x\left[2\left(2x-1\right)^2-3\left(x^2-9\right)-4\left(x+1\right)^2\right]\)
\(=x\left(8x^2-8x+1-3x^2+27-4x^2-8x-4\right)\)
\(=x\left(x^2-16x+28\right)=x\left(x-2\right)\left(x-14\right)\)
\(2x\left(2x-1\right)^2-3x\left(x+3\right)\left(x-3\right)-4x\left(x+1\right)^2\)
\(=2x\left(4x^2-4x+1\right)-3x\left(x^2-9\right)-4x\left(x^2+2x+1\right)\)
\(=8x^3-8x^2+2x-3x^3+27x-4x^3-8x^2-4x\)
\(=x^3-16x^2+25x\)
Rút gọn biểu thức:
\(\left(2x+1\right)^2+\left(3x-1\right)^2+2\left(2x+1\right)\left(3x-1\right)\)
\(\left(2x+1\right)^2+\left(3x-1\right)^2+2\left(2x+1\right)\left(3x-1\right)\))
= \(\left(2x+1\right)^2+2\left(2x+1\right)\left(3x-1\right)+\left(3x-1\right)^2\)
= \(\left[\left(2x+1\right)+\left(3x-1\right)\right]^2\)
= \(\left[2x+1+3x-1\right]^2\)
=\(\left(5x\right)^2\)= \(25x^2\)
Rút gọn các biểu thức sau:
\(a,\left(3x+1\right)^2-2\left(3x+1\right)\left(3x-5\right)+\left(3x-5\right)^2\)
\(b,\left(3x^2-y\right)^2-\left(2x^2+y\right)^2\)
\(a,\left(3x+1\right)^2-2\left(3x+1\right)\left(3x-5\right)+\left(3x-5\right)^2=\left(\left(3x+1\right)-\left(3x-5\right)\right)^2=6^2=36\)
\(b,\left(3x^2-y\right)^2-\left(2x^2+y\right)^2=\left(3x^2-y-2x^2-y\right)\left(3x^2-y+2x^2+y\right)=\left(x^2-2y\right).5x^2\)
a. BT= ((3x+1) - (3x-5))2=62=36
b. BT = (3x2-y-2x2-y). (3x2- y + 2x2+ y) = (x2-2y).5x2
a) \(\left(3x+1\right)^2-2\left(3x+1\right)\left(3x-5\right)+\left(3x-5\right)^2\)
\(=\left[\left(3x+1\right)^2-\left(3x+1\right)\left(3x-5\right)\right]-\left[\left(3x+1\right)\left(3x-5\right)-\left(3x-5\right)^2\right]\)
\(=\left[\left(3x+1\right)\left(3x+1-3x+5\right)\right]-\left[\left(3x+1-3x+5\right)\left(3x-5\right)\right]\)
\(=\left[6\left(3x+1\right)\right]-\left[6\left(3x-5\right)\right]\)
\(=6\left(3x+1-3x+5\right)\)
\(=6.6=36\)