Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)=\(3x^3-15x^2+21x\)
b)\(=-2x^4y-10x^2y+2xy\)
c)\(=-x^3+6x^2+5x-4x^2+24x+20=-x^3+2x^2+29x+20\)
d)\(=2x^4-3x^3+4x^2-2x^2+3x-4=2x^4-3x^32x^2+3x-4\)
e)\(=x^2-4y^2\)
f)\(=-2x^2y^3+y-3\)
g)\(=3xy^4-\dfrac{1}{2}y^2+2x^2y\)
h)\(=9x^2-6x+1-7x^2-14=2x^2-6x-13\)
i)\(=x^2-x-3\)
j)\(=\left(x+2y\right)\left(x^2-2y+4y^2\right):\left(x+2y\right)=x^2-2y+4y^2\)
2: \(\Leftrightarrow\left(x^2+x\right)^2-5\left(x^2+x\right)-6=0\)
\(\Leftrightarrow x^2+x-6=0\)
=>(x+3)(x-2)=0
=>x=-3 hoặc x=2
5: \(\Leftrightarrow\left(x+2\right)\left(x-1\right)\left(x+1\right)=0\)
hay \(x\in\left\{-2;1;-1\right\}\)
a) x3+4x-5 = x3-x2+x2+4x-5=(x3-x2)+(x2-x)+(5x-5)=x2(x-1)+x(x-1)+5(x-1)=(x2+x+5)(x-1)
b) x3-3x2+4=x3-2x2-x2+4=(x3-2x2)-(x2-4)=x2(x-2)-(x-2)(x+2)=(x2-x+2)(x-2)
c) x3+2x2+3x+2=x3+x2+x2+x+2x+2=(x3+x2)+(x2+x)+(2x+2)=x2(x+1)+x(x+1)+2(x+1)=(x2+x+2)(x+1)
d) bạn xem lại đề đúng ko
e) (x2+3x)2-2(x2+3x)-8=x4+6x3+9x2-2x2-6x-8=x4+6x3+7x2-6x-8=x4-x3+7x3-7x2+14x2-14x+8x-8=(x4-x3)+(7x3-7x2)+(14x2-14x)+(8x-8)=x3(x-1)+7x2(x-1)+14x(x-1)+8(x-1)=(x3+7x2+14x+8)(x-1)=(x3+x2+6x2+6x+8x+8)(x-1)=\(\left[\left(x^3+x^2\right)+\left(6x^2+6x\right)+\left(8x+8\right)\right]\left(x-1\right)\)\(=\left[x^2\left(x+1\right)+6x\left(x+1\right)+8\left(x+1\right)\right]\left(x-1\right)\)\(=\left(x^2+6x+8\right)\left(x+1\right)\left(x-1\right)\)\(=\left(x^2+2x+4x+8\right)\left(x+1\right)\left(x-1\right)\)\(=\left[\left(x^2+2x\right)+\left(4x+8\right)\right]\left(x+1\right)\left(x-1\right)\)\(=\left[x\left(x+2\right)+4\left(x+2\right)\right]\left(x+1\right)\left(x-1\right)\)=\(\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x+4\right)\)
f) (x2+4x+10)2-7(x2+4x+11)+7=(x2+4x+10)2-\(\left[7\left(x^2+4x+11\right)-7\right]\)\(=\left(x^2+4x+10\right)^2-7\left(x^2+4x+10\right)\)\(=\left(x^2+4x+10\right)\left(x^2+4x+3\right)\)
a) Ta có: \(x^3+4x-5\)
\(=x^3-x+5x-5\)
\(=x\left(x-1\right)\left(x+1\right)+5\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+5\right)\)
b) Ta có: \(x^3-3x^2+4\)
\(=x^3+x^2-4x^2+4\)
\(=x^2\left(x+1\right)-4\left(x-1\right)\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-4x+4\right)\)
\(=\left(x+1\right)\cdot\left(x-2\right)^2\)
c) Ta có: \(x^3+2x^2+3x+2\)
\(=x^3+x^2+x^2+x+2x+2\)
\(=x^2\left(x+1\right)+x\left(x+1\right)+2\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+x+2\right)\)
d) Ta có: \(x^2+2xy+y^2+2x+2y-3\)
\(=\left(x+y\right)^2+2\left(x+y\right)-3\)
\(=\left(x+y\right)^2+3\left(x+y\right)-\left(x+y\right)-3\)
\(=\left(x+y\right)\left(x+y+3\right)-\left(x+y+3\right)\)
\(=\left(x+y+3\right)\left(x+y-1\right)\)
\(A=6x^2+23x+21-\left(6x^2+23x-55\right)=76\\ B=x^4+x^3-x^2-2x^2-2x+2-x^4-x^3+3x^2+2x\\ =2\\ C=x^4+x^3-3x^2-2x-\left(x^4+x^3-x^2-2x^2-2x+2\right)\\ =-2\)
\(a,=\left(x-2\right)^2\\ b,=\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\\ c,=\left(1-2x\right)\left(1+2x+4x^2\right)\\ d,=\left(x+1\right)^3\\ e,Sửa:\left(x+y\right)^2-9x^2=\left(x+y-3x\right)\left(x+y+3x\right)\\ =\left(y-2x\right)\left(4x+y\right)\\ f,=\left(x+3\right)^2\\ g,=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\\ h,=8\left(x^3-\dfrac{1}{64}\right)=8\left(x-\dfrac{1}{4}\right)\left(x^2+\dfrac{1}{4}x+\dfrac{1}{16}\right)\)
a) \(\left(x-2\right)^2\)
b) \(\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\)
c) \(\left(1-2x\right)\left(1+2x+4x^2\right)\)
d) \(\left(x+1\right)^3\)
e) \(\left(x+y-3\sqrt{x}\right)\left(x+y+3\sqrt{x}\right)\)
f) \(\left(x+3\right)^2\)
g) \(-\left(x-5\right)^2\)
h) \(\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)
4) Ta có: \(\dfrac{2x-5}{5}-\dfrac{x+3}{3}=\dfrac{2-3x}{2}-x-2\)
\(\Leftrightarrow\dfrac{6\left(2x-5\right)}{30}-\dfrac{10\left(x+3\right)}{30}=\dfrac{15\left(2-3x\right)}{30}-\dfrac{30\left(x+2\right)}{30}\)
\(\Leftrightarrow12x-30-10x-30=30-45x-30x-60\)
\(\Leftrightarrow-22x-60=-75x-30\)
\(\Leftrightarrow-22x+75x=-30+60\)
\(\Leftrightarrow53x=30\)
\(\Leftrightarrow x=\dfrac{30}{53}\)
Vậy: \(S=\left\{\dfrac{30}{53}\right\}\)
5) Ta có: \(\dfrac{5x-3}{6}-\dfrac{7x-1}{4}=5\)
\(\Leftrightarrow\dfrac{2\left(5x-3\right)}{12}-\dfrac{3\left(7x-1\right)}{12}=\dfrac{60}{12}\)
\(\Leftrightarrow10x-6-21x+3=60\)
\(\Leftrightarrow-11x-3=60\)
\(\Leftrightarrow-11x=63\)
\(\Leftrightarrow x=-\dfrac{63}{11}\)
Vậy: \(S=\left\{-\dfrac{63}{11}\right\}\)
`9,x^3+x^2-2=0`
`x^3-x^2+2x^2-2=0`
`<=>x^2(x-1)+2(x-1)(x+1)=0`
`<=>(x-1)(x^2+2x+2)=0`
`<=>x=1`
`14,x^2-2x+1=0`
`<=>(x-1)^2=0`
`<=>x-1=0`
`<=>x=1`
`15,x^3+3x^2+3x+1=0`
`<=>(x+1)^3=0`
`<=>x+1=0`
`<=>x=-1`
Bài 6:
1) Ta có: \(2x\left(x-5\right)-\left(x+3\right)^2=3x-x\left(5-x\right)\)
\(\Leftrightarrow2x^2-10x-\left(x^2+6x+9\right)=3x-5x+x^2\)
\(\Leftrightarrow2x^2-10x-x^2-6x-9-3x+5x-x^2=0\)
\(\Leftrightarrow-14x-9=0\)
\(\Leftrightarrow-14x=9\)
\(\Leftrightarrow x=-\dfrac{9}{14}\)
Vậy: \(S=\left\{-\dfrac{9}{14}\right\}\)
`1)2x(x-5)-(x+3)^2=3x-x(5-x)`
`<=>2x^2-10x-x^2-6x-9=3x-5x+x^2`
`<=>x^2-16x-9=x^2-2x`
`<=>14x=-9`
`<=>x=-9/14`
a) Triển khai hằng đẳng thức và rút gọn được 8x + 12 = 0
Từ đó tìm được x = - 3 2
b) Sử dụng hằng đẳng thức, biến đổi phương trình về dạng: (x - 3)(2 x 2 - 4x) = 0
Sưe dụng phương pháp giải PT tích tìm được x ∈ {0; 2; 3}
c) Quy đồng khử mẫu ta được 48x - 16 = 0
Từ đó tìm được x = 1 3
d) Quy đồng khử mẫu ta được 3x + 6 = 2x + 63
Từ đó tìm được x = 57.
