\(\left(3x-2\right)^3=-27\) ae làm hộ cái rồi tích cho
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a)<=> \(\left[\begin{array}{nghiempt}3x-2=7+x\\3x-2=-7-x\end{array}\right.\)
<=> \(\left[\begin{array}{nghiempt}x=\frac{9}{2}\\x=-\frac{5}{4}\end{array}\right.\)
b) | 2x-3|>5<=> \(\left[\begin{array}{nghiempt}2x-3>5\\2x-3< -5\end{array}\right.\)<=>\(\left[\begin{array}{nghiempt}x>4\\x< -1\end{array}\right.\)
c) |3x-1|<7<=>\(\left[\begin{array}{nghiempt}3x-1< 7\\3x-1>-7\end{array}\right.\)<=>\(\left[\begin{array}{nghiempt}x< \frac{8}{3}\\x>-2\end{array}\right.\)
d, xét từng TH1: x<-3/2
TH2:\(\frac{-3}{2}\le0\le\frac{5}{3}\)
TH3:x \(\ge\frac{5}{3}\)
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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) \(x^3+27-x\left(x^2-1\right)=27\)
= \(x^3+27-x^3+x=27\)
= \(27+x=27=>x=0\)
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Đặt \(\hept{\begin{cases}a+b=m\\b+c=n\\c+a=p\end{cases}}\)
Xem VT = A
\(\Rightarrow A=m^2+n^2+p^2-mn-np-mp\)
\(2A=\left(m-n\right)^2+\left(n-p\right)^2+\left(p-m\right)^2\)
\(=\left(a+b-b-c\right)^2+\left(b+c-c-a\right)^2+\left(c+a-a-b\right)^2\)
\(=\left(a-c\right)^2+\left(b-a\right)^2+\left(c-b\right)^2\)
\(=a^2-2ac+c^2+b^2-2ab+a^2+c^2-2bc+b^2\)
\(=2\left(a^2+b^2+c^2-2ab-2bc-2ac\right)\)
\(\Rightarrow A=a^2+b^2+c^2-ab-bc-ca\)(đpcm)
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Đề sai ạ ! Sửa lại nhé :
a) \(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)
\(A=\left(\frac{1}{3}+\frac{3}{x^2-3x}\right):\left(\frac{x^2}{27-3x^2}+\frac{1}{x+3}\right)\)
\(\Leftrightarrow A=\frac{x^2-3x+9}{3\left(x^2-3x\right)}:\left(\frac{-x^2}{3\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}\right)\)
\(\Leftrightarrow A=\frac{x^2-3x+9}{3x\left(x-3\right)}:\frac{-x^2+3\left(x-3\right)}{3\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow A=\frac{x^2-3x+9}{3x\left(x-3\right)}.\frac{3\left(x-3\right)\left(x+3\right)}{-x^2+3x-9}\)
\(\Leftrightarrow A=\frac{-\left(x+3\right)}{x}\)
b) Để \(A\inℤ\)
\(\Leftrightarrow-\left(x+3\right)⋮x\)
\(\Leftrightarrow-x-3⋮x\)
\(\Leftrightarrow3⋮x\)
\(\Leftrightarrow x\inƯ\left(3\right)\)
Vậy để \(A\inℤ\Leftrightarrow x\inƯ\left(3\right)\)(\(x\neℤ\))
Bạn sửa cho mik dòng cuối :
\(x\ne Z\)thành \(x\notin Z\)nhé !
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a) (x3 – 7x + 3 – x2) : (x – 3)
b) (2x4 – 3x2 – 3x2 – 2 + 6x) : (x2 – 2)
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(3x - 2)^ 3= -27
(3x-2)^ 3= (-3)^3
3x-2=-3
3x=-3+2
3x =-1
x= 3 / -1
x = -3 xong rồi nhá