Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(x^3-4x^2-x+4=0\)
=>\(\left(x^3-4x^2\right)-\left(x-4\right)=0\)
=>\(x^2\left(x-4\right)-\left(x-4\right)=0\)
=>\(\left(x-4\right)\left(x^2-1\right)=0\)
=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)
b: Sửa đề: \(x^3+3x^2+3x+1=0\)
=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)
=>\(\left(x+1\right)^3=0\)
=>x+1=0
=>x=-1
c: \(x^3+3x^2-4x-12=0\)
=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)
=>\(x^2\cdot\left(x+3\right)-4\left(x+3\right)=0\)
=>\(\left(x+3\right)\left(x^2-4\right)=0\)
=>\(\left(x+3\right)\left(x-2\right)\left(x+2\right)=0\)
=>\(\left[{}\begin{matrix}x+3=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\\x=-2\end{matrix}\right.\)
d: \(\left(x-2\right)^2-4x+8=0\)
=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)
=>\(\left(x-2\right)^2-4\left(x-2\right)=0\)
=>\(\left(x-2\right)\left(x-2-4\right)=0\)
=>(x-2)(x-6)=0
=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)
a: \(\dfrac{x+10}{4x-8}\cdot\dfrac{4-2x}{x+2}\)
\(=\dfrac{x+10}{4\left(x-2\right)}\cdot\dfrac{-2\left(x-2\right)}{x+2}=\dfrac{-\left(x+10\right)}{2\left(x+2\right)}\)
b: \(\dfrac{1-4x^2}{x^2+4x}:\dfrac{2-4x}{3x}\)
\(=\dfrac{\left(2x-1\right)\left(2x+1\right)}{x\left(x+4\right)}\cdot\dfrac{3x}{2\left(x-2\right)}\)
\(=\dfrac{3\left(2x-1\right)\left(2x+1\right)}{2\left(x-2\right)\left(x+4\right)}\)
c: \(=\dfrac{4y^2}{7x^4}\cdot\dfrac{35x^2}{-8y}=\dfrac{5}{x^2}\cdot\dfrac{-1}{2}\cdot y=\dfrac{-5y}{2x^2}\)
d: \(=\dfrac{\left(x-2\right)\left(x+2\right)}{3\left(x+4\right)}\cdot\dfrac{x+4}{2\left(x-2\right)}=\dfrac{x+2}{6}\)
Chỉ có mỗi biểu thức thôi mà không có yêu cầu thì không thể giải quyết bạn nhé.
<=> x^2^2 + 4x^2 + 8^2 + 2.x^2.4x + 2.x^2.8+ 2.4x.8 + 3x.x^2 +3x.4x = 3x.8 + 2x^2
<=> x^4 + 8x^2 + ..............
Rồi bạn tự tính tiếp nhé
Phân tích đa thức thành nhân tử ?
Ta có: \(P=\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2\)
Đặt \(x^2+4x+8=y\)
Khi đó:
\(P=y^2+3xy+2x^2\)
\(P=\left(y^2+xy\right)+\left(2xy+2x^2\right)\)
\(P=y\left(x+y\right)+2x\left(x+y\right)\)
\(P=\left(x+y\right)\left(2x+y\right)\)
\(P=\left(x^2+5x+8\right)\left(x^2+6x+8\right)\)
\(P=\left(x+2\right)\left(x+4\right)\left(x^2+5x+8\right)\)
1: Ta có: \(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=4x+17\)
\(\Leftrightarrow x^2+6x+9-x^2+4-4x=17\)
\(\Leftrightarrow x=2\)
3: Ta có: \(\left(2x+3\right)\left(x-1\right)+\left(2x-3\right)\left(1-x\right)=0\)
\(\Leftrightarrow2x^2-2x+3x-3+2x-2x^2-3+3x=0\)
\(\Leftrightarrow6x=6\)
hay x=1
a) Ta có: \(x^2+3x-10=0\)
\(\Leftrightarrow x^2+5x-2x-10=0\)
\(\Leftrightarrow x\left(x+5\right)-2\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
Vậy: S={-5;2}
b) Ta có: \(3x^2-7x+1=0\)
\(\Leftrightarrow3\left(x^2-\dfrac{7}{3}x+\dfrac{1}{3}\right)=0\)
mà 3>0
nên \(x^2-\dfrac{7}{3}x+\dfrac{1}{3}=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}-\dfrac{37}{36}=0\)
\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2=\dfrac{37}{36}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{7}{6}=\dfrac{\sqrt{37}}{6}\\x-\dfrac{7}{6}=-\dfrac{\sqrt{37}}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{37}+7}{6}\\x=\dfrac{-\sqrt{37}+7}{6}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{\sqrt{37}+7}{6};\dfrac{-\sqrt{37}+7}{6}\right\}\)
c) Ta có: \(3x^2-7x+8=0\)
\(\Leftrightarrow3\left(x^2-\dfrac{7}{3}x+\dfrac{8}{3}\right)=0\)
mà 3>0
nên \(x^2-\dfrac{7}{3}x+\dfrac{8}{3}=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}+\dfrac{47}{36}=0\)
\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2=-\dfrac{47}{36}\)(vô lý)
Vậy: \(x\in\varnothing\)
hướng dẫn cách làm-tự làm tiếp nha :)
a) đặt \(k=x^2-4x\), ta có:\(k^2-2k=15\)\(\Rightarrow k^2-2x+1=16\Rightarrow\left(k-1\right)^2=4^2=\left(-4\right)^2\)
b) đặt \(A=x^2-3x\), ta có: \(A^2-2A-8=0\Rightarrow A^2-2A+1=9\Rightarrow\left(A-1\right)^2=3^2=\left(-3\right)^2\)
c)theo đề \(\Leftrightarrow\orbr{\begin{cases}x^2-4x+3=0\\x^2-8x+9=0\end{cases}}\)
\(x^2-4x+3=0\Leftrightarrow x^2-4x+4=1\Leftrightarrow\left(x-2\right)^2=1^2=\left(-1\right)^2\)
\(x^2-8x+9=0\Leftrightarrow x^2-8x+16=7\Leftrightarrow\left(x-4\right)^2=\pm\sqrt{7}^2\)
vt ko chi tiết bn ib là đc rùi, sai tớ làm gì T.T
mà tớ làm mẫu 1 bài thui nha, bài còn lại có cách làm òi. bn tự dựa vô nha
\(\text{Đặt }k=x^2-4x,\text{ta có:}\)
\(\left(x^2-4x\right)^2-2.\left(x^2-4x\right)=15\)
\(\Leftrightarrow k^2-2k=0\)
\(\Leftrightarrow k^2-2k+1=16\)
\(\Leftrightarrow\left(k-1\right)^2=16\)
\(\Leftrightarrow\orbr{\begin{cases}k-1=4\\k-1=-4\end{cases}\Leftrightarrow\orbr{\begin{cases}k=5\\k=-3\end{cases}}}\)
\(\text{Với }k=5,\text{Ta có: }x^2-4x=5\Rightarrow x^2-4x-5=0\Rightarrow x^2-5x+x-5=0\)
\(\Rightarrow x.\left(x-5\right)+\left(x-5\right)=0\Rightarrow\left(x+1\right).\left(x-5\right)=0\Rightarrow\orbr{\begin{cases}x=-1\\x=5\end{cases}}\)
\(\text{Với }k=-3,\text{ta có: }x^2-4x=-3\Rightarrow x^2-4x+3=0\Rightarrow k^2-3x-x+3=0\)
\(\Rightarrow x.\left(x-3\right)-\left(x-3\right)=0\Rightarrow\left(x-1\right).\left(x-3\right)=0\Rightarrow\orbr{\begin{cases}x=1\\x=3\end{cases}}\)
Vậy...