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a) \(\left(9x-1\right)^2+\left(1-5x\right)^2+2\left(9x-1\right)\left(1-5x\right)\)
\(=\left(9x-1\right)^2+2\left(9x-1\right)\left(1-5x\right)+\left(1-5x\right)^2\)
\(=\left(9x-1+1-5x\right)^2=\left(4x\right)^2\)
b) \(x^2\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\left(x^2-1\right)\)
\(=x^2\left(x^2-16\right)-\left(x^4-1\right)\)
\(=x^4-16x^2-x^4+1=-16x^2+1\)
a)Ta có:
\(\left(x-2\right)^2-\left(x-3\right)\left(x-1\right)\\ =x^2-4x+4-x^2+4x-3\\ =1\)
Vậy biểu thức \(\left(x-2\right)^2-\left(x-3\right)\left(x-1\right)\)không phụ thuộc vào biến
b) Ta có:
\(\left(x-1\right)^3-\left(x+1\right)^3+6\left(x+1\right)\left(x-1\right)\\ =x^3-3x^2+3x-1-x^3-3x^2-3x-1+6x^2-6\\ =-8\)
Vậy.....
c) Ta có:
\(\left(x-3\right)\left(x+3\right)\left(x^2+9\right)-\left(x^2-2\right)\left(x^2+2\right)\\ =\left(x^2-9\right)\left(x^2+9\right)-x^4+4\\ =x^4-81-x^4+4=-77\)
Vậy....
d) Ta có: \(\left(3x+1\right)^2-2\left(3x+1\right)\left(3x-5\right)+\left(3x-5\right)^2\\ =\left(3x+1-3x+5\right)^2\\ =6^2=36\)
Vậy....
Bài 1:
a,\(3x\left(5x^2-2x-1\right)\)
\(=3x.5x^2-3x.2x-3x=15x^3-6x^2-3x\)
b,\(\left(x^2+2xy-3\right)\left(-xy\right)\)
\(=x^2.\left(-xy\right)+2xy.\left(-xy\right)-3.\left(-xy\right)\)
\(=-x^3y-2x^2y^2+3xy\)
c,\(\dfrac{1}{2}x^2y\left(2x^3-\dfrac{2}{5}xy^2-1\right)\)
\(=\dfrac{1}{2}x^2y.\left(2x^3\right)-\dfrac{1}{2}x^2y.\dfrac{2}{5}xy^2-\dfrac{1}{2}x^2y\)
\(=x^5y-\dfrac{1}{5}x^3y^3-\dfrac{1}{2}x^2y\)
Chúc bạn học tốt!!!
Bài 1:
a) \(3x\left(5x^2-2x-1\right)\\ =15x^3-6x^2-3x\)
b) \(\left(x^2+2xy-3\right)\left(-xy\right)\\ =-x^3y-2x^2y+3xy\)
c) \(\dfrac{1}{2}x^2y\left(2x^3-\dfrac{2}{5}xy^2-1\right)\\ =x^5y-\dfrac{1}{5}x^3y^3-\dfrac{1}{2}x^2y\)
- a, [x^2.(x-3)-(x-3)] :( x-3) = (x-3 ).(x^2-1) : (x-3) =X^2-1
2 b, (x-y-z)^5-3 = (x-y-z)^2
3 c, x^2-1
4 d, 2x^4 + x^2 - 6x^2 + x^3 - 3 - 3x / x^2 - 3
= x^2(2x^2 + x + 1) - 3(2x^2 + x + 1) / x^2 - 3
= (2x^2 + x + 1)(x^2 - 3) / x^2 - 3
= 2x^2 + x + 1
5 e, 2.(x-1)
6 f, (2x3 – 5x2 + 6x – 15) : (2x – 5)
=(2x3−5x2)+(6x−15)=(2x3−5x2)+(6x−15)
=x2(2x−5)+3(2x−5)=x2(2x−5)+3(2x−5)
=(x2+3)(2x−5)=(x2+3)(2x−5)
=(2x3−5x2+6x−15):(2x−5)=x2+3
Bài 1:
a) \(\left(a-b^2\right)\left(a+b^2\right)=a^2-b^4\)
b) \(\left(a^2+2a-3\right)\left(a^2+2a+3\right)=\left(a^2+2a\right)^2-9\)
c) \(\left(a^2+2a+3\right)\left(a^2-2a-3\right)=a^2-\left(2a+3\right)^2\)
d) \(\left(a^2-2a+3\right)\left(a^2+2a+3\right)=9-\left(a^2-2a\right)^2\)
e) \(\left(-a^2-2a+3\right)\left(-a^2-2a+3\right)=\left(-a^2-2a+3\right)^2\)
g) \(\left(a^2+2a+3\right)\left(a^2-2a+3\right)=\left(a^2+3\right)^2-4a^2\)
f) \(\left(a^2+2a\right)\left(2a-a^2\right)=4a^2-a^4\)
Bài 2 :
a) \(\left(x+1\right)\left(x^2-x+1\right)=x^3+1\)
b) \(\left(x+y+z\right)^2=\left(x+y+z\right)\left(x+y+z\right)=x^2+xy+xz+yx+y^2+yz+zx+zy+z^2=x^2+2xy+2yz+2xz+y^2+z^2\)
c) \(\left(x-y+z\right)^2=\left(x-y+z\right)\left(x-y+z\right)=x^2-xy+xz-xy+y^2-yz+xz-yz+z^2=x^2+y^2+z^2-2xy+2xz-2yz\)d) \(\left(x-2y\right)\left(x^2+2xy+4y^2\right)=\left(x-2y\right)^3\)
e) \(\left(x-y-z\right)^2=\left(x-y-z\right)\left(x-y-z\right)=x^2-xy-xz-xy+y^2+yz-xz+yz+z^2=x^2-2xy-2xz+2yz+y^2+z^2\)
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 \(S=\left\{\dfrac{13}{25}\right\}\)
a)\(\left(5x-1\right)^2-\left(5x-4\right)\left(5x+4\right)=7\)
\(\Leftrightarrow25x^2-10x+1-25x^2+16=7\)
\(\Leftrightarrow-10x=-10\)
\(\Leftrightarrow x=1\)
b) k hiểu đề
đề cũng là tìm x mà