a) x2 + 0.36 = 1
b) 25 - (x-1/2)2 =9
c) (2.x -1)2 - (x + 3)2 =0
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26 tháng 9 2021
\(a,=x^3+3x^2+3x+1-x^3+4x^2-4x-1\\ =7x^2-x\\ b,=\left[\left(x+1\right)\left(x^2-x+1\right)\right]\left[\left(x-1\right)\left(x^2+x+1\right)\right]\\ =\left(x^3+1\right)\left(x^3-1\right)=x^6-1\\ c,=8a^2+12a^2+6a+1\\ d,=27a^3-54a^2b+36ab^2-8b^3\)
28 tháng 10 2023
a: \(5^{\left(x-2\right)\left(x+3\right)}=1\)
=>\(\left(x-2\right)\left(x+3\right)=0\)
=>\(\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
c: \(\left|x^2+2x\right|+\left|y^2-9\right|=0\)
mà \(\left\{{}\begin{matrix}\left|x^2+2x\right|>=0\forall x\\\left|y^2-9\right|>=0\forall y\end{matrix}\right.\)
nên \(\left\{{}\begin{matrix}x^2+2x=0\\y^2-9=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\left(x+2\right)=0\\\left(y-3\right)\left(y+3\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\in\left\{0;-2\right\}\\y\in\left\{3;-3\right\}\end{matrix}\right.\)
d: \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=120\)
=>\(2^x\left(1+2+2^2+2^3\right)=120\)
=>\(2^x\cdot15=120\)
=>\(2^x=8\)
=>x=3
e: \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
=>\(\left(x-7\right)^{x+11}-\left(x-7\right)^{x+1}=0\)
=>\(\left(x-7\right)^{x+1}\left[\left(x-7\right)^{10}-1\right]=0\)
=>\(\left[{}\begin{matrix}x-7=0\\x-7=1\\x-7=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\\x=6\end{matrix}\right.\)
5 tháng 4 2022
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)+3\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x+2+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
12 tháng 11 2023
a: \(x^2=2\)
=>\(x^2=\left(\sqrt{2}\right)^2\)
=>\(x=\pm\sqrt{2}\)
b: \(x^2=9\)
=>\(x^2=3^2\)
=>\(\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
c: \(\left(x-\sqrt{2}\right)^2=2\)
=>\(\left[{}\begin{matrix}x-\sqrt{2}=\sqrt{2}\\x-\sqrt{2}=-\sqrt{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\sqrt{2}\\x=0\end{matrix}\right.\)
d: \(4x^2-1=0\)
=>\(4x^2=1\)
=>\(x^2=\dfrac{1}{4}\)
=>\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
14 tháng 11 2023
a, 2\(x\) + 4 - 5\(x\) = -11
-(5\(x\) - 2\(x\)) + 4 = -11
-3\(x\) + 4 = -11
3\(x\) = 11 + 4
3\(x\) = 15
\(x\) = 15 : 3
\(x\) = 5
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
14 tháng 11 2023
b, \(x\) - (-5) = 8
\(x\) + 5 = 8
\(x\) = 8 - 5
\(x\) = 3
30 tháng 9 2016
a) \(\left(2x-1\right)^2-25=0\)
\(\left(2x-1\right)^2=0+25=25\)
\(\left(2x-1\right)^2=5^2=\left(-5\right)^2\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x-1=5\\2x-1=-5\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}2x=6\\2x=-4\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=3\\x=-2\end{array}\right.\)
b) \(8x^3-50x=0\)
\(2x\left(4x^2-25\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x=0\\4x^2-25=0\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=0\\4x^2=25\Rightarrow x^2=\frac{25}{4}\Rightarrow\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-\frac{5}{2}\end{array}\right.\end{array}\right.\)
2 tháng 3 2020
\(3\left(x-1\right)^2-x^2+1=0\)
\(\Leftrightarrow3\left(x-1\right)^2-\left(x^2-1\right)=0\)
\(\Leftrightarrow3\left(x-1\right)^2-\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[3\left(x-1\right)-\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-3-x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\2x=4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=2\end{cases}}}\)
a)x2+0.36=1 b)25-(x-1/2)2=9 c) (2.x-1)2-(x+3)2=0
=> x2=1 25-(x-0.5)2=9 =>(2x-1)2=(x+3)2
x=1 (x-0.5)2=25-9=16 X=-0.666666666666666666
Mà 16=42
x=4+0.5=4.5