(2x+1)+(4x+2)+(6x+3)+...+(20x+10)=275
1. 9x^2 + 12x + 5 = 11
2. 6x^2 + 16x + 12 = 2x^2
3. 16x^2 + 22x + 11 = 6x + 5
4. 12x^2 + 20x + 10 = 3x^2 - 4x
giúp mình với ạ
chuyển vế sang r phân tích thành nhân tử, có thể dùng máy tính bỏ túi nhé bạn
câu 1: 9\(x^2\) + 12\(x\) + 5 =11
(3\(x\))2 + 2.3.\(x\) .2 + 22 + 1 = 11
(3\(x\) + 2)2 = 11 - 1
(3\(x\) + 2)2 = 10
\(\left[{}\begin{matrix}3x+2=\sqrt{10}\\3x+2=-\sqrt{10}\end{matrix}\right.\)
\(\left[{}\begin{matrix}3x=\sqrt{10}-2\\3x=-\sqrt{10}-2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{\sqrt{10}-2}{3}\\x=\dfrac{-\sqrt{10}-2}{3}\end{matrix}\right.\)
Vậy S = {\(\dfrac{-\sqrt{10}-2}{3}\); \(\dfrac{\sqrt{10}-2}{3}\)}
Câu 2: 6\(x^2\) + 16\(x\) + 12 = 2\(x^2\)
6\(x^2\) + 16\(x\) + 12 - 2\(x^2\) = 0
4\(x^2\) + 16\(x\) + 12 = 0
(2\(x\))2 + 2.2.\(x\).4 + 16 - 4 = 0
(2\(x\) + 4)2 = 4
\(\left[{}\begin{matrix}2x+4=2\\2x+4=-2\end{matrix}\right.\)
\(\left[{}\begin{matrix}2x=-2\\2x=-6\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)
S = { -3; -1}
3, 16\(x^2\) + 22\(x\) + 11 = 6\(x\) + 5
16\(x^2\) + 22\(x\) - 6\(x\) + 11 - 5 = 0
16\(x^2\) + 16\(x\) + 6 = 0
(4\(x\))2 + 2.4.\(x\) . 2 + 22 + 2 = 0
(4\(x\) + 2)2 + 2 = 0 (1)
Vì (4\(x\)+ 2)2 ≥ 0 ∀ ⇒ (4\(x\) + 2)2 + 2 > 0 ∀ \(x\) vậy (1) Vô nghiệm
S = \(\varnothing\)
Câu 4. 12\(x^2\) + 20\(x\) + 10 = 3\(x^2\) - 4\(x\)
12\(x^2\) + 20\(x\) + 10 - 3\(x^2\) + 4\(x\) = 0
9\(x^2\) + 24\(x\) + 10 = 0
(3\(x\))2 + 2.3.\(x\).4 + 16 - 6 = 0
(3\(x\) + 4)2 = 6
\(\left[{}\begin{matrix}3x+4=\sqrt{6}\\3x+4=-\sqrt{6}\end{matrix}\right.\)
\(\left[{}\begin{matrix}3x=-4+\sqrt{6}\\3x=-4-\sqrt{6}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{\sqrt{6}-4}{3}\\x=-\dfrac{\sqrt{6}+4}{3}\end{matrix}\right.\)
S = {\(\dfrac{-\sqrt{6}-4}{3}\); \(\dfrac{\sqrt{6}-4}{3}\)}
1)6x-8=3x+1
2)12-10x=25-30x
3)3(2x+3)-2(4x-5)=10x+21
4)5(5x-3)-3(2x-4)11-5x
5)4(2-3x)-5(1-2x)=4-6x
6)8(4x-3)-3(2-3x)=13-40x
7)10x-5(1-4x)=5x-11
8)-3(3-4x)-5(4-3x)=12x-50
9)-2(20x-3)-3(4x-5)=9-2(2x-3)
10)-5(2-3x)+3(5-2x)=3x+3(3-5x)
1)6x-8=3x+1
6x-3x=1+8
3x=9
x=3
Vậy x=3
2: 12-10x=25-30x
=>20x=13
=>x=13/20
3: \(3\left(2x+3\right)-2\left(4x-5\right)=10x+21\)
=>6x+9-8x+10=10x+21
=>10x+21=-2x+19
=>12x=-2
=>x=-1/6
4: \(\Leftrightarrow25x-15-6x+12=11-5x\)
=>19x-3=11-5x
=>24x=14
=>x=7/12
5: \(\Leftrightarrow8-12x-5+10x=4-6x\)
=>4-6x=-2x+3
=>-4x=-1
=>x=1/4
6: \(\Leftrightarrow32x-24-6+9x=13-40x\)
=>41x-30=13-40x
=>81x=43
=>x=43/81
7: \(\Leftrightarrow10x-5+20x=5x-11\)
=>30x-5=5x-11
=>25x=-6
=>x=-6/25
bài3 thức hiện phép chia
a) (2x^3-6x^2+5x-1):(x+1)
b) (4x^3-13x^3+18x^2+20x-5):(x^2-4x+2)
c) (6x^3 -2x^2-9x +3): (3x-1)
a, (2x + 3)^2 + (2x-3)^2 -2 (4x^2 -9) b,(x+3)^3 + (x-2)^3 + x^3 -3x (x+2) (x-2) bài 2:tính a,(6x^2 +y^2) (y^2 - 6x^2) b, (4x + 5) (16x^2 - 20x + 25)
a: Ta có: \(\left(2x+3\right)^2+\left(2x-3\right)^2-2\left(4x^2-9\right)\)
\(=4x^2+12x+9+4x^2-12x+9-8x^2+18\)
\(=36\)
Bài 2:
a: \(\left(y^2+6x^2\right)\left(y^2-6x^2\right)=y^4-36x^4\)
b: \(\left(4x+5\right)\left(16x^2-20x+25\right)=\left(16x^2-25\right)\left(4x-5\right)\)
\(=64x^3-16x^2-100x+125\)
2) 3x^2 + 3x - 6 ; 4) 6x^2 - 13x + 6 ;
5) 6x^2 + 13x + 6 ; 6) 6x^2 + 15x + 6 ;
7) 6x^2 - 15x + 6 ; 8) 6x^2 + 20x + 6 ;
9) 6x^2 - 20x + 6 ; 10) 6x^2 + 12x + 6 ;
11) 8x^2 - 2x - 3 ; 12) 8x^2 + 2x - 3 ;
13) -8x^2 + 5x + 3 ; 14) 8x^2 - 10x - 3 ;
15) 8x^2 + 10x - 3 ; 16) -8x^2 + 23x + 3 ;
17) 8x^2 - 23x - 3 ; 18) 10x^2 - 11x - 6 ;
19) -10x^2 + 11x + 6 ; 20) 10x^2 - 4x - 6 ;
HELP ME!!!
Mik quên mất ghi đề bài r ! Xin lỗi nhé ! Đề bài là:
Bài 2: Phân tích thành nhân tử ( bằng kĩ thuật tách hạng tử).
Đây là toàn bộ nội dung câu hỏi các bạn nhé!
Phân tích đa thức thành nhân tử A. 4x^2-12xy+9y^2-8x+12y B. 3x^2+20x-7 C. (3x-1)^4+2(9y^2-6x+1)+1 D. 2x^3-3x^2+2x-1
a: =(2x-3y)^2-4(2x-3y)
=(2x-3y)(2x-3y-4)
b: =3x^2+21x-x-7
=(x+7)(3x-1)
c: =(3x-1)^4+2(3x-1)^2+1
=[(3x-1)^2+1]^2
d: =2x^3-2x^2-x^2+x+x-1
=(x-1)(2x^2-x+1)
Tìm x, biết:
1) 2x . (x-5) -x . (2x-4) = 15
2) (x+1) . (x+2) - (x+4) . (x+3) = 6
3) 4x2 - 4x+5 - x . (4x-3) = 1-2x
4) (x+3) . (2x+1) - 2x2 = 4x-5
5) -4 . (2x-8) + (2x-1) . (4x+3) = 0
6) -3 . (x-2) + 4 . (2x-6) - 7 . (x-9)= 5 . (3-2)
7) (x-2) . (x+2) -2 . (x-4) = 10. 3x
8) 15x . (x-2) - (5x-1) . (3x + 1) = 6
9) (2x+4) . (x-3) - x . (2x-10) =15-20x
10) (4x-2) . (3x+4) - (2x-1) . (6x+5) = 100
HEPL ME !!! Cần làm gấp những bài này, ai lm dc mk tick cho ng đó nha !!! THANK YOU !!!!
Tìm x, biết:
1) 2x ( x - 5) - x ( 2x - 4 ) = 15
<=> 2x2 - 10x - 2x2 + 4x - 15 = 0
<=> -6x - 15 = 0
<=> -6x = 15
<=> x = -15/6
2) ( x +1)( x + 2 ) - ( x + 4 ) ( x + 3 ) = 6
<=> x2 + 2x + x + 2 - x2 - 3x - 4x - 12 - 6 = 0
<=> -4x = -16
<=> x = 4
3) 4x2 - 4x + 5 - x ( 4x - 3) = 1 - 2x
<=> 4x2 - 4x + 5 - 4x2 + 3x - 1 + 2x = 0
<=> x + 4 = 0
<=> x = -4
4) ( x + 3 ) ( 2x + 1 ) - 2x2 = 4x - 5
<=> 2x2 + x + 6x + 3 - 2x2 - 4x + 5 = 0
<=> 3x + 8 = 0
<=> 3x = -8
<=> x = -8/3
5) -4 ( 2x - 8 ) + ( 2x - 1 )( 4x + 3 ) = 0
<=> - 8x + 32 + 8x2 + 6x - 4x - 3 = 0
.......
6) -3 . (x-2) + 4 . (2x-6) - 7 . (x-9)= 5 . (3-2)
<=> -3x + 6 + 8x - 24 - 7x + 63 - 5 = 0
<=> -2x + 40 = 0
<=> -2x = -40
<=> x = 20
Còn lại tương tự ....
giải các phân thức sau:
a, -3x^2+7x+10=0
b,(x-20x=2x(x+5)
c,(2x-5)(x+11)=(5-2x)(2x=1)
d,x^2+6x+9=4x^2
a. -3x^2+7x+10=0
⇔ 3x^2-7x-10=0
⇔ 3x^2+3x-10x-10=0
⇔ 3x(x+1)-10(x+1)=0
⇔ (3x-10)(x+1)=0
⇔ 3x-10=0 hoặc x+1=0
⇔ x=10/3 hặc x=-1
1)Tìm x biết:
a. 4x . (2x2 - 1) + 27 = (4x2 + 6x + 9) . (2x + 2)
b. (2x + 1) . (5x - 1) = 20x2 - 16x - 1
a: \(\Leftrightarrow8x^3-4x+27=8x^3+8x^2+12x^2+12x+18x+18\)
\(\Leftrightarrow8x^3+20x^2+30x+18=8x^3-4x+27\)
\(\Leftrightarrow20x^2+34x-9=0\)
hay \(x\in\left\{\dfrac{-17+\sqrt{469}}{20};\dfrac{-17-\sqrt{469}}{20}\right\}\)
b: \(\Leftrightarrow20x^2-16x-1=10x^2-2x+5x-1=10x^2+3x-1\)
\(\Leftrightarrow10x^2-19x=0\)
=>x=0 hoặc x=19/10