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\(a.\:\left(2x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49\\ 4x^2+12x+9-4x^2+4=49\\ 12x=49-9\\ x=\dfrac{40}{12}=\dfrac{10}{3}\)

12 tháng 7 2017

4x2 +12x+9-4(x2-1)=49

4x2+12x+9-4x+4=49

12x+13=49

12x=36

x=3

12 tháng 7 2017

a, Xem lại đề:

b, \(16x^2-\left(4x-5\right)^2=15\)

\(\Rightarrow16x^2-\left(16x^2-40x+25\right)=15\)

\(\Rightarrow16x^2-16x^2+40x-25=15\)

\(\Rightarrow40x=40\Rightarrow x=1\)

Chúc bạn học tốt!!!

\(a.\:\left(7x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49\\ 49x^2+42x+9-4x^2+4=49\\ 45x^2+42x+13=49\\ x^2+\dfrac{42}{45}x+\dfrac{13}{45}=\dfrac{49}{45}\\ x^2+2.\dfrac{7}{15}x+\left(\dfrac{7}{15}\right)^2=\dfrac{49}{45}-\dfrac{13}{45}+\left(\dfrac{7}{15}\right)^2\\ \left(x+\dfrac{7}{15}\right)^2=\dfrac{229}{225}\\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{7}{15}=\dfrac{229}{225}\\x+\dfrac{7}{15}=-\dfrac{229}{225}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{124}{225}\\x=-\dfrac{334}{225}\end{matrix}\right.\)

\(c.\:\left(3x+4\right)^2-\left(3x+1\right)\left(3x-1\right)\\ =9x^2+24x+16-9x^2+1\\ 40x=-1\\ x=-\dfrac{1}{40}\)

\(d.\:\left(3x-1\right)^2-\left(3x-2\right)^2=0\\ \left(3x-1+3x-2\right)\left(3x-1-3x+2\right)=0\\ \left(6x-3\right)=0\\ x=\dfrac{1}{2}\)

\(g.\:\left(2x+1\right)^2-\left(x-1\right)^2=0\\ \left(2x+1+x-1\right)\left(2x+1-x+1\right)=0\\ 3x\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

12 tháng 7 2017

c,\(\left(3x+4\right)^2-\left(3x-1\right)\left(3x+1\right)=49\)

\(\Rightarrow9x^2+24x+16-\left(9x^2-1\right)=49\)

\(\Rightarrow9x^2+24x+16-9x^2+1=49\)

\(\Rightarrow24x=49-1-16\)

\(\Rightarrow24x=32\Rightarrow x=\dfrac{4}{3}\)

d, \(\left(3x-1\right)^2-\left(3x-2\right)^2=0\)

\(\Rightarrow\left(3x-1-3x+2\right).\left(3x-1+3x-2\right)=0\)

\(\Rightarrow6x-3=0\Rightarrow6x=3\Rightarrow x=\dfrac{1}{2}\)

e, \(\left(2x+1\right)^2-\left(x-1\right)^2=0\)

\(\Rightarrow\left(2x+1-x+1\right)\left(2x+1+x-1\right)=0\)

\(\Rightarrow\left(x+2\right).3x=0\Rightarrow x.\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

Chúc bạn học tốt!!!

18 tháng 7 2017

a) \(\left(4x-1\right)^2-\left(3x+2\right)\left(3x-2\right)=\left(7x-1\right)\left(x+2\right)+\left(2x+1\right)^2-\left(4x^2+7\right)\)(1)

\(\Leftrightarrow\left(16x^2-8x+1\right)-\left(9x^2-4\right)=\left(7x^2+14x-x-2\right)+\left(4x^2+4x+1\right)-\left(4x^2+7\right)\)

\(\Leftrightarrow16x^2-8x+1-9x^2+4=7x^2+13x-2+4x^2+4x+1-4x^2-7\)

\(\Leftrightarrow7x^2-8x+5=7x^2+17x-8\)

\(\Leftrightarrow7x^2-8x-7x^2-17x=-8-5\)

\(\Leftrightarrow-25x=-13\)

\(\Leftrightarrow x=\dfrac{13}{25}\)

Vậy tập nghiệm phương trình (1) là \(S=\left\{\dfrac{13}{25}\right\}\)

18 tháng 7 2017

gắp cái gì

a: Đặt \(C=3\left(x-1\right)^2-\left(x+1\right)^2+2\left(x-3\right)\left(x+3\right)-\left(2x+3\right)^2-\left(5-20x\right)\)

\(D=5x\left(x-7\right)\left(x+7\right)-x\left(2x-1\right)^2-\left(x^3+4x^2-246x\right)-175\)

Do đó: A=C+D

\(C=3\left(x-1\right)^2-\left(x+1\right)^2+2\left(x-3\right)\left(x+3\right)-\left(2x+3\right)^2-\left(5-20x\right)\)

\(=3x^2-6x+3-x^2-2x-1+2x^2-18-\left(4x^2+12x+9\right)-5+20x\)

\(=4x^2-8x-16-4x^2-12x-9-5+20x\)

\(=-30\)

\(D=5x\left(x-7\right)\left(x+7\right)-x\left(2x-1\right)^2-\left(x^3+4x^2-246x\right)-175\)

\(=5x\left(x^2-49\right)-x\left(4x^2-4x+1\right)-x^3-4x^2+246x-175\)

\(=5x^3-245x-4x^3+4x^2-x-x^3-4x^2+246x-175\)

=-175

A=C+D=-30-175=-205

b: Đặt \(E=-2x\left(3x+2\right)^2+\left(4x+1\right)^2+2\left(x^3+8x^2+3x-2\right)-\left(5-x\right)\)

\(F=\left(5x-2\right)^2-\left(6x+1\right)^2+11\left(x-2\right)\left(x+2\right)-16\left(3-2x\right)\)

Do đó: B=E+F

\(E=-2x\left(3x+2\right)^2+\left(4x+1\right)^2+2\left(x^3+8x^2+3x-2\right)-\left(5-x\right)\)

\(=-2x\left(9x^2+12x+4\right)+16x^2+8x+1+2x^3+16x^2+6x-4-5+x\)

\(=-18x^3-24x^2-8x+32x^2+14x+1-5+x\)

\(=-18x^3+8x^2+7x-4\)

\(F=\left(5x-2\right)^2-\left(6x+1\right)^2+11\left(x-2\right)\left(x+2\right)-16\left(3-2x\right)\)

\(=25x^2-20x+4-36x^2-12x-1+11x^2-44-48+32x\)

\(=-95\)

\(B=-18x^3+8x^2+7x-99\)

4 tháng 7 2017

Ta có : (a + b)(a2 - ab + b2) - 2a(a - b)2

= (a + b).(a - b)2  - 2a(a - b)2

= (a - b)2(a + b - 2a)

a: \(A=\dfrac{3x^2+4x^2y}{x^2}-\dfrac{10xy+15xy^2}{5y}\)

\(=3+4y-2x-3xy\)

\(=3+4\cdot\left(-5\right)-2\cdot2-3\cdot2\cdot\left(-5\right)\)

\(=3-20-4+30=10-1=9\)

b: \(B=\dfrac{18a^4-27a^3}{9a^2}-10a^3:5a\)

\(=2a^2-3a-10a^3:5a\)

\(=2a^2-3a-2a^2=-3a=-3\cdot\left(-8\right)=24\)

c: \(C=\dfrac{8x^3-4x^2}{2x^2}-\dfrac{4x^2-3x}{x}+2x\)

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

=2x+1=-2+1=-1

9 tháng 9 2017

a)A=88x2-132x+198+12x2-18x+27-8x3+2

A=100x2-150x+227-8x3

b)B=x2-2x+1+(-4x2-4).(x3+x2+x-x2-x-1)

B=x2-2x+1-4x5-4x4-4x3+4x4+4x3+4x2-4x3-4x2-4x+4x2+4x+1

B=5x2-2x+2-4x5-4x3

10 tháng 9 2017

a) A=(x-2)(x+2)(x+3)-(x+1)2

= x2 - 22. ( x + 3) - (x+1) 2
= x2 - 4x - 12 - x2 - 2x.1 + 12
= x2 - 4x - 12 - x2 - 2x + 1
= -16x - 11

10 tháng 9 2017

b) B=(x+2)3-(x-2)3-12x2
= x3 + 3x2.2 + 3x.22 + 23 - x3 + 3x2.2 + 3x.22 + 23 - 12x2
= -12x2


14 tháng 8 2020

a) 16x^2 - (4x - 5)^2 = 15

<=> 16x^2 - 16x^2 + 40x - 25 = 15

<=> 40x = 40

<=> x = 1

b) (2x + 3)^2 - 4(x - 1)(x + 1) = 49

<=> 4x^2 + 12x + 9 - 4x^2 - 4x + 4x + 4 = 49

<=> 12x + 13 = 49

<=> 12x = 36

<=> x = 3

c) (2x + 1)(1 - 2x) + (1 - 2x)^2 = 18

<=> 1 - 4x^2 + 1 - 4x + 4x^2 = 18

<=> 2 - 4x = 18

<=> -4x = 16

<=> x = -4

d)2(x + 1)^2 - (x - 3)(x + 3) - (x - 4)^2 = 0

<=> 2x^2 + 4x + 2 - x^2 + 3^2 - x^2 + 8x - 16 = 0

<=> 12x - 5 = 0

<=> 12x = 5

<=> x = 5/12

e) (x - 5)^2 - x(x - 4) = 9

<=> x^2 - 10x + 25 - x^2 + 4x = 9

<=> -6x + 25 = 9

<=> -6x = 9 - 25

<=> -6x = -16

<=> x = -16/-6 = 8/3

f) (x - 5)^2 + (x - 4)(1 - x) = 0

<=> x^2 - 10x + 25 + x - x^2 - x - 4 + 4x = 0

<=> -5x + 21 = 0

<=> -5x = -21

<=> x = 21/5