Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
câu a là hằng đẳng thức luôn
A=(2x+4)^2
B khai triển tung tóe ra thì phần sau triệt tiêu hết còn 4(a^2+b^2+c^2)
câu c cảm giác sai đề vì mấy câu này phải là (3x)^ ms ra hdt chứ nhỉ
\(A=\left(a^2+b^2-c^2\right)^2-\left(a^2-b^2+c^2\right)^2-4a^2b^2\)
\(=\left(a^2+b^2-c^2+a^2-b^2+c^2\right)\left(a^2+b^2-c^2-a^2+b^2-c^2\right)-4a^2b^2\)
\(=2a^2.2b^2-4a^2b^2=0\)
\(C=\left(2-6x\right)^2+\left(2-5x\right)^2+2\left(6x-2\right)\left(2-5x\right)\)
\(=\left[\left(2-6x\right)+\left(2-5x\right)\right]^2\)
\(=\left[4-11x\right]^2\)
\(=16-88x+121x^2\)
chúc bn học tốt
a) ( x - 5 )( 2x + 3 ) + 2x( 1 - x )
= 2x2 - 7x - 15 + 2x - 2x2
= -5x - 15
= -5( x + 3 )
b) ( 3x - 5 )2 - ( x + 5 )( 5 - x ) - 5/2( -2x )2
= 9x2 - 30x + 25 + ( x + 5 )( x - 5 ) - 5/2.4x2
= 9x2 - 30x + 25 + x2 - 25 - 10x2
= -30x
c) ( 3x + 2 )( 4 - 6x + 9x2 ) - 3x( 3x - 2 )2 + 12( -2/3 - 3x2 )
= ( 3x )3 + 23 - 3x( 9x2 - 12x + 4 ) - 8 - 36x2
= 27x3 + 8 - 27x3 + 36x2 - 12x - 8 - 36x2
= -12x
\(a,\left(3x+1\right)^2-2\left(3x+1\right)\left(3x-5\right)+\left(3x-5\right)^2=\left(\left(3x+1\right)-\left(3x-5\right)\right)^2=6^2=36\)
\(b,\left(3x^2-y\right)^2-\left(2x^2+y\right)^2=\left(3x^2-y-2x^2-y\right)\left(3x^2-y+2x^2+y\right)=\left(x^2-2y\right).5x^2\)
a. BT= ((3x+1) - (3x-5))2=62=36
b. BT = (3x2-y-2x2-y). (3x2- y + 2x2+ y) = (x2-2y).5x2
rút gọn biểu thức
a)2x(2x−1)2−3x(x+3)(x−3)−4x(x+1)2
=2x(4x2-4x+1)-3x.(x2-9)-4x(x2+2x+1)
=8x3-8x2+2x-3x3-27x-4x3-8x2-4x
=8x3-16x2-7x3-29x
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: \(B=\left(3x+2\right)^2+2\left(3x+2\right)\left(1-2y\right)+\left(2y-1\right)^2\)
\(=\left(3x+2+1-2y\right)^2\)
\(=\left(3x-2y+3\right)^2\)
Câu A) là \(\left(2x+y\right)^2-\left(y-2x\right)^2\)
Chứ ko phải là\(\left(2x+y\right)^2-\left(2x-y\right)^2\)
Nhưng dù sao thì cũng cảm ơn
a: Ta có: \(\left(3x-1\right)^2-2\left(5x-2\right)^2-2\left(x^2+x-1\right)\left(x-1\right)\)
\(=9x^2-6x+1-2\left(25x^2-20x+4\right)-2\left(x^3-x^2+x^2-x-x+1\right)\)
\(=9x^2-6x+1-50x^2+40x-8-2\left(x^3-2x+1\right)\)
\(=-41x^2+34x-7-2x^3+4x-2\)
\(=-2x^3-41x^2+38x-9\)
b: Ta có: \(\left(3a+1\right)^2+2\left(9a^2-1\right)+\left(3a-1\right)^2\)
\(=\left(3a+1+3a-1\right)^2\)
\(=36a^2\)
Bài 1:
a)(4x-3)(3x+2)-(6x+1)(2x-5)+1
=12x2-x-6-12x2+28x+5+1
=27x
b)(3x+4)2+(4x-1)2+(2+5x)(2-5x)
=9x2+24x+16+16x2-8x+1+4-25x2
=16x+21
c)(2x+1)(4x2-2x+1)+(2-3x)(4+6x+9x2)-9
=8x3+1+8-27x3-9
=-19x3
Bài 2:
a)3x(x-4)-x(5+3x)=-34
=>3x2-12x-3x2-5x=-34
=>-17x=-34
=>x=2
Vậy x=2
b)(3x+1)2+(5x-2)2=34(x+2)(x-2)
=>9x2+6x+1+25x2-20x+4=34(x2-4)
=>34x2-14x+5-34x2+136=0
=>-14x+141=0
=>-14x=-141
=>x=\(\frac{141}{14}\)
Vậy x=\(\frac{141}{14}\)
c)x3+3x2+3x+28=0
=>x3-x2+7x+4x2-4x+28=0
=>x(x2-x+7)+4(x2-x+7)=0
=>(x+4)(x2-x+7)=0
\(\Rightarrow\left[\begin{array}{nghiempt}x+4=0\\x^2-x+7=0\left(2\right)\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=-4\\\left(2\right)\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{27}{4}>0\end{array}\right.\)
=>(2) vô nghiệm
Vậy x=-4
\(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)=2^{16}-1\)
\(B=\left(3x-1\right)^2+\left(5-3x\right)^2+\left(6x-2\right)\left(5-3x\right)\)
\(=\left(3x-1\right)^2+\left(5-3x\right)^2+2.\left(3x-1\right)\left(5-3x\right)\)
\(=\left(3x-1+5-3x\right)^2=4^2=16\)