tìm x bết
I 2x - 1 I = (-4)^2
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ND
4
XO
15 tháng 8 2020
6x(3x + 5) - 2x(3x - 2) + (17 - x)(x - 1) + x(x - 18) = 0
=> (18x2 - 6x2 - x2 + x2) + (30x + 4x - 16x - 18x) - 17 = 0
=> 12x2 - 17 = 0
=> 12x2 = 17
=> x2 = 17/12
=> \(\orbr{\begin{cases}x=\sqrt{\frac{17}{12}}\\x=-\sqrt{\frac{17}{12}}\end{cases}}\)
TP
0
28 tháng 10 2023
a: \(\left(\dfrac{1}{4}-x\right)\left(x+\dfrac{2}{5}\right)=0\)
=>\(\left[{}\begin{matrix}\dfrac{1}{4}-x=0\\x+\dfrac{2}{5}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{2}{5}\end{matrix}\right.\)
b: \(\left|2x+1\right|+\dfrac{3}{2}=2\)
=>\(\left|2x+1\right|=\dfrac{1}{2}\)
=>\(\left[{}\begin{matrix}2x+1=\dfrac{1}{2}\\2x+1=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{1}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
c: (2x-3)2=36
=>\(\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
d: \(7^{x+2}+2\cdot7^x=357\)
=>\(7^x\cdot49+7^x\cdot2=357\)
=>\(7^x=7\)
=>x=1
T
28 tháng 10 2023
a) \(\left(\dfrac{1}{4}-x\right)\left(x+\dfrac{2}{5}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{4}-x=0\\x+\dfrac{2}{5}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{2}{5}\end{matrix}\right.\)
\(---\)
b) \(\left|2x+1\right| +\dfrac{2}{3}=2\)
\( \Rightarrow\left|2x+1\right|=2-\dfrac{2}{3}\)
\(\Rightarrow\left|2x+1\right|=\dfrac{4}{3}\)
\(\Rightarrow\left[{}\begin{matrix}2x+1=\dfrac{4}{3}\\2x+1=-\dfrac{4}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{1}{3}\\2x=-\dfrac{7}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=-\dfrac{7}{6}\end{matrix}\right.\)
\(---\)
c) \(\left(2x-3\right)^2=36\)
\(\Rightarrow\left(2x-3\right)^2=\left(\pm6\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(---\)
d) \(7^{x+2}+2\cdot7^x=357\)
\(\Rightarrow7^x\cdot7^2+2\cdot7^x=357\)
\(\Rightarrow7^x\cdot\left(7^2+2\right)=357\)
\(\Rightarrow7^x\cdot\left(49+2\right)=357\)
\(\Rightarrow7^x\cdot51=357\)
\(\Rightarrow7^x=357:51\)
\(\Rightarrow7^x=7\)
\(\Rightarrow x=1\)
LK
2
TT
25 tháng 10 2020
1.
a) \(=3x\left(x-3y\right)\left(12x+1\right)\)
b) \(=\left(x-1\right)^2-4y^2=\left(x-2y-1\right)\left(x+2y-1\right)\)
c) \(=y\left(x-z\right)+3\left(x-z\right)=\left(x-z\right)\left(y+3\right)\)
d) \(=\left(x-1\right)^2-49=\left(x-8\right)\left(x+6\right)\)
e) Chịu!
f) Chịu!
3.
a) \(\Leftrightarrow\left(x-5\right)^2-5\left(x-5\right)=0\)
⇔ \(\left(x-5\right)\left(x-10\right)=0\)
⇔ \(\left[{}\begin{matrix}x=5\\x=10\end{matrix}\right.\)
b) \(A=x\left(5x-3y\right)-z\left(3y-5x\right)=\left(x+z\right)\left(5x-3y\right)\)
Thay x = 2015,2016; y = 1,4; = -2015,2016 vào A ta có
A = 0
25 tháng 10 2020
1,
a, =36x^2.(x-3y)+3x.(x-3y)
=(36x^2+3x)(x-3y)=3x(12x+1)(x-3y)
b, =(x^2-2x+1)-4y^2
=(x-1)^2-4y^2
=(x-1-4y)(x-1+4y)
C, =xy+3x-yz-3z
=x(y+3)-z(y+3)=(x-z)(y+3)
d, x^2-2x+1-49
=(x-1)^2-7^2
e, chịu
f, chịu
2, a, chịu
b, =x(5x-3y)+z(5x-3y)
=(x+z)(5x-3y), thay vào ta có
=(2015,2016+-2015,2016)(5.2015,2016-3.-2015,2016)
=0
Mấy bài chịu để suy nghĩ đã sorry
|2x - 1| = (-4)2
=> |2x - 1| = 16
TH1: 2x - 1 = 16
=> 2x = 16 + 1
=> 2x = 17
=> x = 17 : 2
=> x = 8,5
TH2: 2x - 1 = -16
=> 2x = -16 + 1
=> 2x = -15
=> x = -15 : 2
=> x = -7,5
Vậy x = 8,5 hoặc x = -7,5
|2x-1|=(-4)2
<=> |2x-1|=16
=> 2x-1=16 hoặc 2x-1=-16
• 2x-1=16 => x=17/2
• 2x-1=-16 => x=-15/2
Vậy ...