Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
c) \(\left(x+1\right)^4+\left(x^2+x+1\right)^2\)
\(=\left(x+1\right)^4+x^4+x^2+1+2x^3+2x^2+2x\)
\(=\left(x+1\right)^4+x^4+3x^2+1+2x^3+2x\)
a) \(x^4-7x^3+14x^2-7x+1\)(1)
Giả sử x khác 0, khi đó :
\(\left(1\right)\Leftrightarrow x^2\left(x^2-7x+14-\dfrac{7}{x}+\dfrac{1}{x^2}\right)\)
\(\Leftrightarrow x^2\left[\left(x^2+\dfrac{1}{x^2}\right)-7\left(x+\dfrac{1}{x}\right)+14\right]\)
\(\Leftrightarrow x^2\left[\left(x^2+2\cdot x\cdot\dfrac{1}{x}+\dfrac{2}{x^2}\right)-2-7\left(x+\dfrac{1}{x}\right)+14\right]\)
\(\Leftrightarrow x^2\left[\left(x+\dfrac{1}{x}\right)^2-7\left(x+\dfrac{1}{x}\right)+12\right]\)
Đặt \(x+\dfrac{1}{x}=a\)
pt \(\Leftrightarrow x^2\left(a^2-7a+12\right)\)
\(\Leftrightarrow x^2\left(a^2-3a-4a+12\right)\)
\(\Leftrightarrow x^2\left[a\left(a-3\right)-4\left(a-3\right)\right]\)
\(\Leftrightarrow x^2\left(a-3\right)\left(a-4\right)\)
\(\Leftrightarrow x^2\left(x+\dfrac{1}{x}-3\right)\left(x+\dfrac{1}{x}-4\right)\)
a: \(4\left(x+5\right)\left(x+6\right)\left(x+10\right)\left(x+12\right)-3x^2\)
\(=4\left(x^2+60+17x\right)\left(x^2+60+16x\right)-3x^2\)
\(=\left(2x^2+120+34x\right)\left(2x^2+120+32x\right)-3x^2\)
\(=\left(2x^2+120\right)^2+66x\left(2x^2+120\right)+1085x^2\)
\(=\left(2x^2+120\right)^2+31x\left(2x^2+120\right)+35x\left(2x^2+120\right)+1085x^2\)
\(=\left(2x^2+120\right)\left(2x^2+31x+120\right)+35x\left(2x^2+120+31x\right)\)
\(=\left(2x^2+31x+120\right)\left(2x^2+35x+120\right)\)
b: \(x^4-8x+63\)
\(=x^4+4x^3+9x^2-4x^3-16x^2-36x+7x^2+28x+63\)
\(=\left(x^2+4x+9\right)\left(x^2-4x+7\right)\)
mk ghi đáp án, còn lại bạn tự biến đổi
a) \(2x^3-x^2+5x+3=\left(2x+1\right)\left(x^2-x+3\right)\)
b) \(x^3+5x^2+8x+4=\left(x+1\right)\left(x+2\right)^2\)
c) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
d) \(4x^4+1=\left(2x^2-2x+1\right)\left(2x^2+2x+1\right)\)
e) \(x^4-7x^3+14x^2-7x+1=\left(x^2-4x+1\right)\left(x^2-3x+1\right)\)
mk làm chi tiết theo yêu của của người hỏi đề:
a) \(2x^3-x^2+5x+3\)
\(=\left(2x^3-2x^2+6x\right)+\left(x^2-x+3\right)\)
\(=2x\left(x^2-x+3\right)+\left(x^2-x+3\right)\)
\(=\left(2x+1\right)\left(x^2-x+3\right)\)
b) \(x^3+5x^2+8x+4\)
\(=\left(x^3+4x^2+4x\right)+\left(x^2+4x+4\right)\)
\(=x\left(x^2+4x+4\right)+\left(x^2+4x+4\right)\)
\(=\left(x+1\right)\left(x^2+4x+4\right)\)
\(=\left(x+1\right)\left(x+2\right)^2\)
\(a,\left(2x^2+1\right)+4x>2x\left(x-2\right)\)
\(\Leftrightarrow2x^2+1+4x>2x^2-4x\)
\(\Leftrightarrow4x+4x>-1\)
\(\Leftrightarrow8x>-1\)
\(\Leftrightarrow x>-\frac{1}{8}\)
\(b,\left(4x+3\right)\left(x-1\right)< 6x^2-x+1\)
\(\Leftrightarrow4x^2-4x+3x-3< 6x^2-x+1\)
\(\Leftrightarrow4x^2-x-3< 6x^2-x+1\)
\(\Leftrightarrow4x^2-6x^2< 1+3\)
\(\Leftrightarrow-2x^2< 4\)
\(\Leftrightarrow x^2>2\)
\(\Leftrightarrow x>\pm\sqrt{2}\)
a) ĐKXĐ: x khác +2
\(\frac{x-2}{2+x}-\frac{3}{x-2}-\frac{2\left(x-11\right)}{x^2-4}\)
<=> \(\frac{x-2}{2+x}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}\)
<=> (x - 2)^2 - 3(2 + x) = 2(x - 11)
<=> x^2 - 4x + 4 - 6 - 3x = 2x - 22
<=> x^2 - 7x - 2 = 2x - 22
<=> x^2 - 7x - 2 - 2x + 22 = 0
<=> x^2 - 9x + 20 = 0
<=> (x - 4)(x - 5) = 0
<=> x - 4 = 0 hoặc x - 5 = 0
<=> x = 4 hoặc x = 5
làm nốt đi
a)3x2+4x-9x-12=0
=>(3x2+4x)-(9x+12)=0
=> x(3x+4)-3(3x+4)=0
=> (x-3)(3x+4)=0 =>x-3=0 hoặc 3x+4=0
=>tự tính
b)7x2-9x+2=0
=>7x2-7x-2x+2=0
=>(7x2-7x)-(2x-2)=0
=>7x(x-1)-2(x-1)=0
=>(7x-2)(x-1)=0
=>như câu a
bạn chỉ biết làm 2 câu thôi
Bài 1:
a: \(A=\dfrac{x^4+x^3+x+1}{x^4-x^3+2x^2-x+1}=\dfrac{x^3\left(x+1\right)+\left(x+1\right)}{x^4-x^3+x^2+x^2-x+1}\)
\(=\dfrac{\left(x+1\right)\left(x^3+1\right)}{\left(x^2-x+1\right)\left(x^2+1\right)}=\dfrac{\left(x+1\right)^2}{x^2+1}\)
Để A=0 thì x+1=0
hay x=-1
b: \(B=\dfrac{x^4-5x^2+4}{x^4-10x^2+9}=\dfrac{\left(x^2-1\right)\left(x^2-4\right)}{\left(x^2-1\right)\left(x^2-9\right)}=\dfrac{x^2-4}{x^2-9}\)
Để B=0 thi (x-2)(x+2)=0
=>x=2 hoặc x=-2
a) \(3x^2-5x-12=0\)
\(\Leftrightarrow3x^2+4x-9x-12=0\)
\(\Leftrightarrow x\left(3x+4\right)-3\left(3x+4\right)=0\)
\(\Leftrightarrow\left(3x+4\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x+4=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{4}{3}\\x=3\end{cases}}\)
b) \(7x^2-9x+2=0\)
\(\Leftrightarrow7x^2-7x-2x+2=0\)
\(\Leftrightarrow7x\left(x-1\right)-2\left(x-1\right)=0\).
\(\Leftrightarrow\left(7x-2\right)\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}7x-2=0\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{2}{7}\\x=1\end{cases}}\)
Đề có ghi gì đâu mà bất định với có định :VV
"Dùng phương pháp hệ số bất định" để làm gì?