Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(x-3\right)\left(x^2+3x+9\right)+x\left(5-x^2\right)=6x\)
\(\Leftrightarrow x^3+3x^2+9x-3x^2-9x-27+5x-x^3-6x=0\)
\(\Leftrightarrow-x=27\)
\(\Leftrightarrow x=-27\)
a.\(\Leftrightarrow\left(x-1\right)^3+8-x^3+3x\left(x+2\right)=17\)
\(\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=17\)
\(\Leftrightarrow9x+7=17\)
\(\Leftrightarrow9x=10\Leftrightarrow x=\frac{10}{9}\)
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...
\(1,\left(2-x\right)^2-9=0\)
\(\Leftrightarrow\left(2-x-9\right)\left(2-x+9\right)=0\)
\(\Leftrightarrow\left(-7-x\right)\left(11-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-7-x=0\\11-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=11\end{matrix}\right.\)
\(b,\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)=15-9\left(x+1\right)^2\)\(\Leftrightarrow x^3-9x^2+27x-27-x^3-27=15-9x^2-18x-9\)\(\Leftrightarrow x^3-9x^2+27x-x^3+9x^2+18x=15+27+27\)\(\Leftrightarrow45x=69\Rightarrow x=\dfrac{23}{15}\)
1. \(\left(2-x\right)^2-9=0\)
\(\left(2-x\right)^2=9\)
\(\left(2-x\right)^2=3^2\)
\(2-x=3\)
\(-x=-1\Rightarrow x=1\)
giải
bài 2;tìm x
a)(x+2)(x^2-2x+4)-x(x^2+2)=15
x^3-2x^2+4x+2x^2-4x+8-x^3-2x=15
-2x+8=15
-2x=15-8
-2x=7
x=-7/2
vậy x=-7/2
\(a,\left(3x+x\right)\left(x^2-9\right)-\left(x-3\right)\left(x^2+3x+9\right)\)
\(=4x\left(x^2-9\right)-x^3+27\)
\(=4x^3-36x-x^3+27\)
\(=3x^3-36x+27\)
\(\left(x+6\right)^2-2x.\left(x+6\right)+\left(x-6\right).\left(x+6\right)\)
\(=\left(x+6\right).\left(x+6-2x+x-6\right)\)
\(=\left(x+6\right).0\)
\(=0\)
\(\left(x-5\right)\left(x+5\right)-\left(x+3\right)^2+3\left(x-2\right)^2=\left(x+1\right)^2-\left(x-4\right)\left(x+4\right)+3x^2\)\(\Leftrightarrow x^2-25-\left(x^2+6x+9\right)+3\left(x^2-4x+4\right)=x^2+2x+1-\left(x^2-4^2\right)+3x^2\)\(\Leftrightarrow x^2-25-x^2-6x-9+3x^2-12x+12=x^2+2x+1-x^2+16+3x^2\)
\(\Leftrightarrow-20x=39\)
\(\Leftrightarrow x=\frac{-39}{20}\)
Vậy \(x=\frac{-39}{20}\)
Bài giải
\(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=15\)
\(x^3-3x^2+9x+3x^2-9x+27-x\left(x^2-2^2\right)=15\)
\(x^3+27-x^3+2^2x=15\)
\(27-4x=15\)
\(4x=12\)
\(x=3\)
x bằng 3 nha