Tìm X:
a) (x + 3)2 – x2 + 15 = 1
b) (5 – x)2 + 6x – x2 = –7
c) (1 + x)(1 – x) – (x – 4)2 = –15
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TP
3
4 tháng 9 2021
a: Ta có: \(2x^3-18x=0\)
\(\Leftrightarrow2x\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
b: Ta có: \(\left(3x-2\right)\left(2x+1\right)-6x\left(x+2\right)=11\)
\(\Leftrightarrow6x^2+3x-4x-2-6x^2-12x=11\)
\(\Leftrightarrow-13x=13\)
hay x=-1
c: Ta có: \(\left(x-1\right)^3-\left(x+2\right)\left(x^2-2x+4\right)=3\left(1-x^2\right)\)
\(\Leftrightarrow x^3-3x^2+3x-1-x^3-8=3-3x^2\)
\(\Leftrightarrow3x=12\)
hay x=4
4 tháng 9 2021
a) 2x3-18x=0
⇔ 2x(x2-9)=0
⇔ 2x(x-3)(x+3)=0
⇔ \(\left\{{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
b)(3x-1)(2x+1)-6x(x+2)=11
⇔ 6x2+x-1-6x2-12x=11
⇔ -11x=12
\(\Leftrightarrow x=-\dfrac{12}{11}\)
c) (x-1)3-(x+2).(x2-2x+4)=3.(1-x2)
⇔ x3-3x2+3x-1-x3-8-3+3x2=0
⇔ 3x=12
⇔ x=4
C
3
7 tháng 8 2021
a) Ta có: \(36x^3-4x=0\)
\(\Leftrightarrow4x\left(9x^2-1\right)=0\)
\(\Leftrightarrow x\left(3x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=\dfrac{-1}{3}\end{matrix}\right.\)
b) Ta có: \(3x\left(x-2\right)+x-2=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-1}{3}\end{matrix}\right.\)
TI
1
10 tháng 10 2021
\(4\left(x+1\right)\left(x^2-x+1\right)-x^3+3x=15\)
\(\Leftrightarrow4x^3+4-x^3+3x-15=0\)
\(\Leftrightarrow3x^3+3x-11=0\)
Đây là phương trình vô tỉ nhé bạn, nên nghiệm rất xấu!
13 tháng 11 2021
\(1,\Leftrightarrow x^2+10x+25=x^2-4x-21\\ \Leftrightarrow14x=-46\\ \Leftrightarrow x=-\dfrac{23}{7}\\ 2,\Leftrightarrow x^3+8=15+x^3+2x\\ \Leftrightarrow2x=-7\Leftrightarrow x=-\dfrac{7}{2}\\ 3,\Leftrightarrow\left(x+3\right)^2=0\\ \Leftrightarrow x=-3\\ 4,\Leftrightarrow x^3-9x^2+27x-27=0\\ \Leftrightarrow\left(x-3\right)^3=0\\ \Leftrightarrow x-3=0\Leftrightarrow x=3\\ 5,\Leftrightarrow4x^2+4x+1-4x^2-16x-16=9\\ \Leftrightarrow-12x=24\Leftrightarrow x=-2\\ 6,\Leftrightarrow x^2-3x+5x-15=0\\ \Leftrightarrow\left(x-3\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
14 tháng 9 2021
a) \(4x+9=0\Leftrightarrow4x=-9\Leftrightarrow x=-\dfrac{9}{4}\)
b) \(-5x+6=0\Leftrightarrow5x=6\Leftrightarrow x=\dfrac{6}{5}\)
c) \(x^2-1=0\Leftrightarrow\left(x-1\right)\left(x+1\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
d) \(x^2-9=0\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
e) \(x^2-x=0\Leftrightarrow x\left(x-1\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
f) \(x^2-2x=0\Leftrightarrow x\left(x-2\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
g) \(\left(x-4\right)\left(x^2+1\right)=0\Leftrightarrow x-4=0\Leftrightarrow x=4\)( do \(x^2+1\ge1>0\))
h) \(3x^2-4x=0\Leftrightarrow x\left(3x-4\right)=0\Leftrightarrow\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{3}\end{matrix}\right.\)
i) \(x^2+9=0\Leftrightarrow x^2=-9\)( vô lý do \(x^2\ge0>-9\))
Vậy \(x\in\left\{\varnothing\right\}\)
QL
1
26 tháng 1 2022
\(a,x=\dfrac{13}{2}-2\\ x=\dfrac{9}{2}\\ b,x=\dfrac{4}{5}\times\dfrac{3}{4}\\ x=\dfrac{12}{20}=\dfrac{3}{5}\)
HN
1 tháng 7 2021
a)
⇔ \(x^2-16=9\)
⇔ \(x^2=25\)
⇔ \(x=\pm5\)
b)
⇔ \(x^2-4x+4-25x^2+20x-4=0\)
⇔ \(16x-24x^2=0\)
⇔ \(8x\left(2-3x\right)=0\)
⇒ \(\left[{}\begin{matrix}x=0\\2-3x=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy \(x=0\) hoặc \(x=\dfrac{2}{3}\)
c)
⇔ \(3x^2-10x-20=0\)
⇔ \(x^2-2.x.\dfrac{5}{3}+\dfrac{25}{9}-\dfrac{205}{9}=0\)
⇔ \(\left(x-\dfrac{5}{3}\right)^2=\dfrac{205}{9}\)
⇒ \(\left[{}\begin{matrix}x-\dfrac{5}{3}=\sqrt{\dfrac{205}{9}}\\x-\dfrac{5}{3}=-\sqrt{\dfrac{205}{9}}\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=\dfrac{\sqrt{\text{205}}}{\text{3}}+\dfrac{5}{3}\\x=-\dfrac{\sqrt{\text{205}}}{\text{3}}+\dfrac{5}{3}\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=\dfrac{15+\text{9}\sqrt{\text{205}}}{\text{9}}\\\text{x}=-\dfrac{15+\text{9}\sqrt{\text{205}}}{\text{9}}\end{matrix}\right.\)
Vậy...
d)
⇔ \(\left(x^2+x\right)^2-49=\left(x^2+x\right)^2-7x\)
⇔ 7x = 49
⇔ x=7
Vậy...
30 tháng 9 2021
a: Ta có: \(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(\Leftrightarrow6x-9+4-2x=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
a)(x+3)2-x2+15=2x+6-2x+15=1
=21=1
Bạn chép sai đầu bài à
x2 là x^2 à