a, 7x2+4x+1
b, 9x2+12x+8
c, x2+10x+30
d, x2+x+3
a,x3+3x2+3x+1
b,x2+6x+9
c,-x3+9x2-27x+27
d,x2+4x+4
k,10x-25-x2
f,(x+y)2-9x2
g,8x3+42x2y+16xy2+6xy+y3
a) \(x^3+3x^2+3x+1=x^2+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=\left(x-1\right)^3\)
b) \(x^2+6x+9=x^2+2\cdot3\cdot x+3^2=\left(x+3\right)^2\)
c) \(-x^3+9x^2-27x+27\)
\(=-\left(x^3-9x^2+27x-27\right)\)
\(=-\left(x^3-3\cdot3\cdot x^2+3\cdot3^2\cdot x-3^3\right)=-\left(x-3\right)^3\)
d) \(x^2+4x+4=x^2+2\cdot2\cdot x+2^2=\left(x+2\right)^2\)
k) \(10x-25-x^2=-x^2+10x-25=-\left(x^2-10x+25\right)\)
\(=-\left(x^2-2\cdot5\cdot x+5^2\right)=-\left(x-5\right)^2\)
f) \(\left(x+y\right)^2-9x^2=\left(x-y\right)^2-\left(3x\right)^2=\left[\left(x-y\right)-3x\right]\left[\left(x-y\right)+3x\right]\)
\(=\left(x-y-3x\right)\left(x-y+3x\right)=\left(-2x-y\right)\left(4x-y\right)\)
Giải các phương trình tích sau:
1.a)(3x – 2)(4x + 5) = 0 b) (2,3x – 6,9)(0,1x + 2) = 0
c)(4x + 2)(x2 + 1) = 0 d) (2x + 7)(x – 5)(5x + 1) = 0
2. a)(3x + 2)(x2 – 1) = (9x2 – 4)(x + 1)
b)x(x + 3)(x – 3) – (x + 2)(x2 – 2x + 4) = 0
c)2x(x – 3) + 5(x – 3) = 0 d)(3x – 1)(x2 + 2) = (3x – 1)(7x – 10)
3.a)(2x – 5)2 – (x + 2)2 = 0 b)(3x2 + 10x – 8)2 = (5x2 – 2x + 10)2
c)(x2 – 2x + 1) – 4 = 0 d)4x2 + 4x + 1 = x2
4. a) 3x2 + 2x – 1 = 0 b) x2 – 5x + 6 = 0
c) x2 – 3x + 2 = 0 d) 2x2 – 6x + 1 = 0
e) 4x2 – 12x + 5 = 0 f) 2x2 + 5x + 3 = 0
Bài 1:
a) (3x - 2)(4x + 5) = 0
<=> 3x - 2 = 0 hoặc 4x + 5 = 0
<=> 3x = 2 hoặc 4x = -5
<=> x = 2/3 hoặc x = -5/4
b) (2,3x - 6,9)(0,1x + 2) = 0
<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0
<=> 2,3x = 6,9 hoặc 0,1x = -2
<=> x = 3 hoặc x = -20
c) (4x + 2)(x^2 + 1) = 0
<=> 4x + 2 = 0 hoặc x^2 + 1 # 0
<=> 4x = -2
<=> x = -2/4 = -1/2
d) (2x + 7)(x - 5)(5x + 1) = 0
<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0
<=> 2x = -7 hoặc x = 5 hoặc 5x = -1
<=> x = -7/2 hoặc x = 5 hoặc x = -1/5
bài 2:
a, (3x+2)(x^2-1)=(9x^2-4)(x+1)
(3x+2)(x-1)(x+1)=(3x-2)(3x+2)(x+1)
(3x+2)(x-1)(x+1)-(3x-2)(3x+2)(x+1)=0
(3x+2)(x+1)(1-2x)=0
b, x(x+3)(x-3)-(x-2)(x^2-2x+4)=0
x(x^2-9)-(x^3+8)=0
x^3-9x-x^3-8=0
-9x-8=0
tự tìm x nha
a) x(4x+3y)−(y−2x)2
b) (3+x)(x−3)−(x−1)(x2−3)
c)−2(x−3)2+(x+1)(5x−1)
d) (2x+1)(4x2−2x+1)−3x2(x−2)
e) (3x2+19x+20):(3x+4)
f) (7x2+x3+12x−6):(x2+4x−3)
\(a,=4x^2+3xy-y^2+4xy-4x^2=7xy-y^2\\ b,=x^2-9-x^3+3x+x^2-3=-x^3+2x^2+3x-12\\ c,=-2x^2+12x-18+5x^2+4x-1=3x^2+16x-19\\ d,=8x^3+1-3x^3+6x^2=5x^3+6x^2+1\\ e,=\left(3x^2+4x+15x+20\right):\left(3x+4\right)\\ =\left(3x+4\right)\left(x+5\right):\left(3x+4\right)\\ =x+5\\ f,=\left(x^3+4x^2-3x+3x^2+12x-9+3x+3\right):\left(x^2+4x-3\right)\\ =\left[\left(x^2+4x-3\right)\left(x+3\right)+3x+3\right]:\left(x^2+4x-3\right)\\ =x+3\left(dư.3x+3\right)\)
Giải các phương trình sau:
a, (9x2 - 4)(x + 1) = (3x +2)(x2 - 1)
b, (x - 1)2 - 1 + x2 = (1 - x)(x + 3)
c, (x2 - 1)(x + 2)(x - 3) = (x - 1)(x2 - 4)(x + 5)
d, x4 + x3 + x + 1 = 0
e, x3 - 7x + 6 = 0
f, x4 - 4x3 + 12x - 9 = 0
g, x5- 5x3 + 4x = 0
h, x4 - 4x3 + 3x2 + 4x - 4 = 0
a, \(\Leftrightarrow\left(9x^2-4\right)\left(x+1\right)-\left(3x+2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(\left(9x^2-4\right)-\left(\left(3x+2\right)\left(x-1\right)\right)\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(9x^2-4-\left(3x^2-x-2\right)\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(9x^2-4-3x^2+x+2\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x+1\right)=0;3x^2+x-2=0\)
=> x=-1
với \(3x^2+x-2=0\)
ta sử dụng công thức bậc 2 suy ra : \(x=\dfrac{2}{3};x=-1\)
Vậy ghiệm của pt trên \(S\in\left\{-1;\dfrac{2}{3}\right\}\)
b: \(\Leftrightarrow x^2-2x+1-1+x^2=x+3-x^2-3x\)
\(\Leftrightarrow2x^2-2x=-x^2-2x+3\)
\(\Leftrightarrow3x^2=3\)
hay \(x\in\left\{1;-1\right\}\)
c: \(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x-3\right)-\left(x-1\right)\left(x-2\right)\left(x+2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[\left(x+1\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-2x-3-x^2-3x+10\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(-5x+7\right)=0\)
hay \(x\in\left\{1;-2;\dfrac{7}{5}\right\}\)
CMR giá trị của các biểu thức sau không âm với mọi giá trị của biến x: A=x2 –3x+10 B = x2 – 5x + 2021 C = 4x2 + 4x + 5 D = 9x2 – 12x + 6
a: ta có: \(A=x^2-3x+10\)
\(=x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{31}{4}\)
\(=\left(x-\dfrac{3}{2}\right)^2+\dfrac{31}{4}>0\forall x\)
b: Ta có: \(B=x^2-5x+2021\)
\(=x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}+\dfrac{8015}{4}\)
\(=\left(x-\dfrac{5}{2}\right)^2+\dfrac{8015}{4}>0\forall x\)
Tìm giá trị lớn nhất hoặc nhỏ nhất của biểu thức:
a) A = x2-4x+20
b) B = x2+3x+7
c) C = -x2-10x+70
d) D = -4x2+12x+1
\(A=x^2-4x+20=x^2-4x+4+16=\left(x-2\right)^2+16\)
Do \(\left(x-2\right)^2\ge0\)
\(\Rightarrow\left(x-2\right)^2+16\ge16\)
\(\Rightarrow Min\left(A\right)=16\)
\(B=x^2-3x+7=x^2-3x+\dfrac{9}{4}-\dfrac{9}{4}+7=\left(x-\dfrac{3}{2}\right)^2+\dfrac{19}{4}\)
Do \(\left(x-\dfrac{3}{2}\right)^2\ge0\)
\(\Rightarrow\left(x-\dfrac{3}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\)
\(\Rightarrow Min\left(B\right)=\dfrac{19}{4}\)
\(C=-x^2-10x+70=-\left(x^2+10x+25\right)+25+70=-\left(x-5\right)^2+95\)
Do \(-\left(x-5\right)^2\le0\)
\(\Rightarrow-\left(x-5\right)^2+95\le95\)
\(\Rightarrow Max\left(C\right)=95\)
\(D=-4x^2+12x+1=-\left(4x^2-12x+9\right)+9+1=-\left(2x-3\right)^2+10\)
Do \(-\left(2x-3\right)^2\le0\)
\(\Rightarrow-\left(2x-3\right)^2+10\le10\)
\(\Rightarrow Max\left(D\right)=10\)
a) A = -3x(x-5) +3( x2 -4x) -3x-10
b) B = 4x( x2 -7x +2) – 4( x3 -7x2 +2x -5)
c) C = 5x( x2 – x) – x2( 5x-5) -15
d) D = 7( x2 -5x+3)- x( 7x-35) -14
e) E = x2 - 4x - x( x-4) -15
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,A=x2+x-2 b,B=4x-x2+5 c,C=9x2-6x+3 d,D=3x+x2-7 e,E=x2+y2-3x+2y+3 f,F=x2+y2-x+4y+5
Lần sau bạn chú ý viết đầy đủ yêu cầu của đề bài.
a) \(A=x^2+x-2=\left(x+2\right)\left(x-1\right)\)
b) \(B=-x^2+4x+5=-\left(x^2-4x-5\right)=-\left(x-5\right)\left(x+1\right)\)
c) \(C=9x^2-6x+3=3\left(3x^2-2x+1\right)\)
Tìm x, biết
b) x2 - 2x + 1 = 4
c) x2 - 4x + 4 = 9
d) 4x2 - 4x + 1 = 4
e) x2 - 2x - 8 = 0
f) 9x2 - 6x - 8 = 0
b)x2-2x+1=4
⇔(x-1)2=4
\(\Leftrightarrow\left[{}\begin{matrix}x-1=2\\x-1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
c)x2-4x+4=9
⇔ (x-2)2=9
\(\Leftrightarrow\left[{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
d)4x2-4x+1=4
⇔ (2x-1)2=4
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=4\\2x-1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-3}{2}\end{matrix}\right.\)
e)x2-2x-8=0
⇔ x2-4x+2x-8=0
⇔ x(x-4)+2(x-4)=0
⇔(x-4)(x+2)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)
f)9x2-6x-8=0
⇔ 9x2-12x+6x-8=0
⇔ 3x(3x-4)+2(3x-4)=0
⇔ (3x-4)(3x+2)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=\dfrac{-2}{3}\end{matrix}\right.\)
giải phương trình sau:
a. (9x2-4)(x+1) = (3x+2) (x2-1)
b. (x-1)2-1+x2 = (1-x)(x+3)
c. (x2-1)(x+2)(x-3) = (x-1)(x2-4)(x+5)
d. x4+x3+x+1=0
e. x3-7x+6 = 0
f. x4-4x3+12x-9 = 0
g. x5-5x3+4x = 0
h. x4-4x3+3x2+4x-4 = 0
m.n jup vs