help mk vs huhu
tìm x biết
\(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-1\right)\left(x+1\right)=27\)
1) \(27+\left(x-3\right).\left(x^2+3x+9\right)=-x\)
2) \(-4\left(x+2\right)-7\left(2x-1\right)+9\left(4-3x\right)=30\)
3) \(x^2-4x+4=0\)
4) \(\left(x-1\right).\left(x^2+x+1\right)-x.\left(x+2\right).\left(x-2\right)=5\)
Tìm x giúp mk vs nha mai mk đi hk rùi
1)
\(27+(x-3)(x^2+3x+9)=-x\)
\(\Leftrightarrow 27+(x^3-3^3)=-x\)
\(\Leftrightarrow x^3=-x\)
\(\Leftrightarrow x^3+x=0\Leftrightarrow x(x^2+1)=0\)
\(\Rightarrow \left[\begin{matrix} x=0\\ x^2+1=0(vl)\end{matrix}\right.\)
Vậy $x=0$
2)
\(-4(x+2)-7(2x-1)+9(4-3x)=30\)
\(\Leftrightarrow -4x-8-14x+7+36-27x=30\)
\(\Leftrightarrow -45x+35=30\Leftrightarrow -45x=-5\)
\(\Rightarrow x=\frac{-5}{-45}=\frac{1}{9}\)
3)
\(x^2-4x+4=0\)
\(\Leftrightarrow x^2-2.2x+2^2=0\)
\(\Leftrightarrow (x-2)^2=0\Rightarrow x-2=0\Rightarrow x=2\)
4)
\((x-1)(x^2+x+1)-x(x+2)(x-2)=5\)
\(\Leftrightarrow (x^3-1^3)-x[(x+2)(x-2)]=5\)
\(\Leftrightarrow x^3-1-x(x^2-2^2)=5\)
\(\Leftrightarrow x^3-1-x^3+4x=5\)
\(\Leftrightarrow 4x-1=5\Rightarrow 4x=6\Rightarrow x=\frac{6}{4}=\frac{3}{2}\)
tìm x biết
a) \(\left(x-2\right)^3\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)
b)\(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
c)\(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)
d)\(\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)=0\)
Giúp mk vs đc k ạ mk đg cần gấp
\(b,\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Leftrightarrow x^3+8-x^3-2x=15\)
\(\Leftrightarrow-2x=15-8=7\)
\(\Leftrightarrow x=\frac{-7}{2}\)
Vậy \(x=\frac{-7}{2}\)
Tìm x bt
1) \(3.\left(2x-1\right).\left(3x-1\right)-\left(2x-3\right).\left(9x-1\right)-3=-3\)\(-3\)
2) \(3x-1.\left(2x+7\right)-x+1.\left(6x-5\right)=x+2-\left(x-5\right)\)
3) \(3xy.\left(x+y\right)-\left(x+y\right).\left(x^2+y^2+2xy\right)+y^3=27\)
4) \(5.x+1.\left(1-x\right).\left(2x^2+3\right)=0\)
giúp mk vs mai mk đi hk rùi
Tìm x biết:
\(a.3x^2-3x\left(x-2\right)=36\)
\(b.x\left(2x-1\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+4,5\right)=3,5\)
\(c.\left(3x^2-x+1\right)\left(x-1\right)+x^2\left(4-3x\right)=\frac{5}{2}\)
Giúp mk vs ạ <3 <3
a)\(\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow6x=36\Leftrightarrow x=6\)
Tìm x biết :
a ) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=27\)
b ) \(2x^2+7x+3\) = 0
c ) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=27\)
d ) \(2x^2+11x+9=0\)
e ) \(x\left(x+2\right)-x^2-8=0\)
f ) \(\left(x-3\right)\left(x^2+3x+9\right)-x\left(x-2\right)\left(x+2\right)=27\)
a) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=27\)
\(\Rightarrow x^3+3^3-x\left(x^2-4\right)=27\)
\(\Rightarrow x^3+27-x^3+4x=27\)
\(\Rightarrow27+4x=27\)
\(\Rightarrow4x=0\)
\(\Rightarrow x=0\)
b) \(2x^2+7x+3=0\)
\(\Rightarrow2x^2+x+6x+3=0\)
\(\Rightarrow x\left(2x+1\right)+3\left(2x+1\right)=0\)
\(\Rightarrow\left(2x+1\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x+1=0\\x+3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=-1\\x=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-3\end{matrix}\right.\)
c) Trùng đề bài a
d) \(2x^2+11x+9=0\)
\(\Rightarrow2x^2+2x+9x+9=0\)
\(\Rightarrow2x\left(x+1\right)+9\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(2x+9\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\2x+9=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\2x=-9\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{9}{2}\end{matrix}\right.\)
giúp mk vs huhu
bài 1 : tính tích của các đa thức sau
a. \(\left(3x+4x^2-2\right)\left(-x^2+1+2x\right)\)
Bài 2 : thực hiện phép tính
\(12a^2b\left(a-b\right)\left(a+b\right)\)
bài 3 : \(3\left(1-4x\right)\left(x-1\right)+4\left(3x-2\right)\left(x+3\right)=-27\)
Bài 1:
\(\left(3x+4x^2-2\right)\left(-x^2+1+2x\right)\)
\(=-3x^3+3x+6x^2-4x^4+4x^2+8x^3+2x^2-2-4x\)
\(=5x^3-x+12x^2-4x^2-2\)
Bài 2:
\(12a^2b\left(a-b\right)\left(a+b\right)\)
\(=12a^2b\cdot\left(a^2-b^2\right)\)
\(=12a^4b-12a^2b^3\)
Bài 3:
\(3\left(1-4x\right)\left(x-1\right)+4\left(3x-2\right)\left(x+3\right)=-27\) (1)
\(\Leftrightarrow\left(3-12x\right)\left(x-1\right)+\left(12x-8\right)\left(x+3\right)=-27\)
\(\Leftrightarrow3x-3-12x^2+12x+12x^2+36x-8x-24=-27\)
\(\Leftrightarrow43x-27=-27\)
\(\Leftrightarrow43x=0\)
\(\Leftrightarrow x=0\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{0\right\}\)
Tìm x biết:
\(\left(x-1\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+3\left(x^2-4\right)=2\)
(x-1)^3-(x+3)(x^2-3x+9)+3(x^2-4)=2
=>x^3-3x^2+3x-1-x^3-27+3x^2-12=2
=>3x-40=2
=>x=42/3=14
\(\left(3-x\right)^3=-\dfrac{27}{64};\left(x-5\right)^3=\dfrac{1}{-27};\left(x-\dfrac{1}{2}\right)^3=\dfrac{27}{8};\left(2x-1\right)^2=\dfrac{1}{4};\left(2-3x\right)^2=\dfrac{9}{4};\left(1-\dfrac{2}{3}\right)^2=\dfrac{4}{9}\)
\(\left(3-x\right)^3=-\dfrac{27}{64}\)
\(\left(3-x\right)^3=\left(\dfrac{-3}{4}\right)^3\)
\(=>3-x=\dfrac{-3}{4}\)
\(x=3-\dfrac{-3}{4}=\dfrac{12}{4}+\dfrac{3}{4}\)
\(x=\dfrac{15}{4}\)
________
\(\left(x-5\right)^3=\dfrac{1}{-27}\)
\(\left(x-5\right)^3=\left(\dfrac{-1}{3}\right)^3\)
\(=>x-5=\dfrac{-1}{3}\)
\(x=\dfrac{-1}{3}+5=\dfrac{-1}{3}+\dfrac{15}{3}\)
\(x=\dfrac{14}{3}\)
_____________
\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{27}{8}\)
\(\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{3}{2}\right)^3\)
\(=>x-\dfrac{1}{2}=\dfrac{3}{2}\)
\(x=\dfrac{3}{2}+\dfrac{1}{2}\)
\(x=2\)
________
\(\left(2x-1\right)^2=\dfrac{1}{4}\)
\(\left(2x-1\right)^2=\left(\dfrac{1}{2}\right)^2\) hoặc \(\left(2x-1\right)^2=\left(\dfrac{-1}{2}\right)^2\)
\(=>2x-1=\dfrac{1}{2}\) \(2x-1=\dfrac{-1}{2}\)
\(2x=\dfrac{1}{2}+1=\dfrac{1}{2}+\dfrac{2}{2}\) \(2x=\dfrac{-1}{2}+1=\dfrac{-1}{2}+\dfrac{2}{2}\)
\(2x=\dfrac{3}{2}\) \(2x=\dfrac{1}{2}\)
\(x=\dfrac{3}{2}:2=\dfrac{3}{2}.\dfrac{1}{2}\) \(x=\dfrac{1}{2}:2=\dfrac{1}{2}.\dfrac{1}{2}\)
\(x=\dfrac{3}{4}\) \(x=\dfrac{1}{4}\)
____________
\(\left(2-3x\right)^2=\dfrac{9}{4}\)
\(\left(2-3x\right)^2=\left(\dfrac{3}{2}\right)^2\) hoặc \(\left(2-3x\right)^2=\left(\dfrac{-3}{2}\right)^2\)
\(=>2-3x=\dfrac{3}{2}\) \(2-3x=\dfrac{-3}{2}\)
\(3x=2-\dfrac{3}{2}=\dfrac{4}{2}-\dfrac{3}{2}\) \(3x=2-\dfrac{-3}{2}=\dfrac{4}{2}+\dfrac{3}{2}\)
\(3x=\dfrac{1}{2}\) \(3x=\dfrac{7}{2}\)
\(x=\dfrac{1}{2}.\dfrac{1}{3}\) \(x=\dfrac{7}{2}.\dfrac{1}{3}\)
\(x=\dfrac{1}{6}\) \(x=\dfrac{7}{6}\)
______________
\(\left(1-\dfrac{2}{3}\right)^2=\dfrac{4}{9}\) -> Kiểm tra đề câu này
(3-x)3=(-\(\dfrac{3}{4}\))3
3-x=-\(\dfrac{3}{4}\)
x=3-(-\(\dfrac{3}{4}\))
x=\(\dfrac{15}{4}\)
dạng 1: thực hiện phép tính :
a,\(12a^2b\left(a-b\right)\left(a+b\right)\)
b,\(\left(2x^2-3x+5\right)\left(x^2-8x+2\right)\)
DẠNG 2: TÌM x
a,\(\frac{1}{4}x^2-\left(\frac{1}{2}x-4\right)\frac{1}{2}x=-14\)
b,\(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-1\right)\left(x+1\right)=27\)
LÀM ĐƯỢC CÂU NÀO THÌ GIẢI HỘ VS Ạ
a) = \(12a^2b\left(a^2-b^2\right)\)
= \(12a^4b-12a^2b^3\)
b)nhân ra :
= \(2x^4-16x^3+4x^2-3x^3+24x^2-6x+5x^2-40x+10\)
= \(2x^4-19x^3+33x^2-46x+10\)
Tìm x:
a) \(\frac{1}{4}x^2-\left(\frac{1}{4}x^2-2x\right)=-14\)
= \(\frac{1}{4}x^2-\frac{1}{4}x^2+2x=-14\)
=\(2x=-14=>x=-7\)
b) \(x^3+27-x\left(x^2-1\right)=27\)
= \(x^3+27-x^3+x=27\)
= \(27+x=27=>x=0\)