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a)
(2x+1)2=25
=> \(\left[\begin{array}{nghiempt}2x+1=5\\2x+1=-5\end{array}\right.\)
=>\(\left[\begin{array}{nghiempt}2x=4\\2x=-6\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=2\\x=-3\end{array}\right.\)
d)
(x-1)3=-125
=> x-1=-5
=> x=-4
còn câu b và c bạn viết đề rõ hơn nha
b3: Vì x:y:z= a:b:c
nên x/a= y/b=z/c
ADTCCDTSBN, ta có:
x/a=y/b=z/c= (x/a)^2=(y/b)^2=(z/c)^2=(x+y+z)^2
x/a=y/b=z/c suy ra (x/a)^2=(y/b)^2=(z/c)^2=(x+y+z)^2
suy ra x^2/a^2 = y^2/b^2 = z^2/c^2= (x+y+z)^2
ADTCCDTSBN, có:
(x+y+z)^2= x^2/a^2=...=z^2/c^2=x^2+y^2+z^2/a^2+b^2+c^2= x^2+y^2+z^2/1= x^2+y^2+z^2
Vậy...
\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{8}{125}\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{2}{5}\right)^3\Rightarrow x-\dfrac{1}{2}=\dfrac{2}{5}\)
\(\Rightarrow x=\dfrac{2}{5}+\dfrac{1}{2}\Rightarrow x=\dfrac{9}{10}\)
\(2^x+2^{x+3}=144\Rightarrow2^x+2^x.2^3=144\Rightarrow2^x\left(1+2^3\right)=144\Rightarrow9.2^x=144\Rightarrow2^x=144:9=16\Rightarrow2^x=2^4\Rightarrow x=4\)
thank you!
Thật ra câu này mk làm rồi nhưng chưa chắc chắn cho lắm!
A)
\(x^3=-125\)
\(x=-5\)
Vậy x = -5
B)
| x | = x + 1
\(\Leftrightarrow\orbr{\begin{cases}x=x+1\\x=-\left(x+1\right)\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-x=1\\x=-x-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}0=1\left(VN\right)\\2x=-1\end{cases}}\)
\(\Leftrightarrow x=-\frac{1}{2}\)
Vậy x = -1/2
C) \(x^2=x\)
\(\Leftrightarrow x^2-x=0\)
\(\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
Vậy x1 = 0 ; x2 = 1
A) x3 = -125
x = -125 : 3
x \(\approx\)-41,5
B) | x | = x + 1
\(\orbr{\begin{cases}x=x\\x=-x\end{cases}}\)
=>\(\orbr{\begin{cases}x>0\\x< 0\end{cases}}\)
C) x2 = x
x \(\varepsilon\)N
~ Hok tốt ~
Bài 1:
a) Ta có: 2x + 2x+3 = 144
2x.(1+23) = 144
2x.9 = 144
2x = 16
x = 4
a. 2x +2x+3=144 b.3x + 3x+2 = 810
=>2x .1 +2x.23=144 =>3x.1 +3x.32=810
=>2x.(1+8)=144 =>3x.(1+9) =810
=>2x = 16 =>3x = 81
=>2x = 24 => 3x =34
=>x = 4 =>x = 4
vậy x=4 vậy x=4
a) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\left(-\dfrac{7}{12}\right)\cdot1\dfrac{2}{5}\)
\(\Rightarrow\dfrac{1}{6}x=\left(-\dfrac{7}{12}\right)\cdot\dfrac{7}{5}\)
\(\Rightarrow\dfrac{1}{6}x=-\dfrac{49}{60}\)
\(\Rightarrow x=-\dfrac{49}{60}:\dfrac{1}{6}\)
\(\Rightarrow x=-\dfrac{49}{10}\)
b) \(\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\dfrac{9}{4}\)
\(\Rightarrow\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\left(\pm\dfrac{3}{2}\right)^2\)
+) \(\dfrac{1}{5}-\dfrac{3}{2}x=\dfrac{3}{2}\)
\(\Rightarrow\dfrac{3}{2}x=\dfrac{1}{5}-\dfrac{3}{2}\)
\(\Rightarrow\dfrac{3}{2}x=-\dfrac{13}{10}\)
\(\Rightarrow x=-\dfrac{13}{10}:\dfrac{3}{2}\)
\(\Rightarrow x=-\dfrac{13}{15}\)
+) \(\left(1,25-\dfrac{4}{5}x\right)^3=-125\)
\(\Rightarrow\left(\dfrac{5}{4}-\dfrac{4}{5}x\right)^3=\left(-5\right)^3\)
\(\Rightarrow\dfrac{5}{4}-\dfrac{4}{5}x=-5\)
\(\Rightarrow\dfrac{4}{5}x=\dfrac{5}{4}+5\)
\(\Rightarrow\dfrac{4}{5}x=\dfrac{25}{4}\)
\(\Rightarrow x=\dfrac{25}{4}:\dfrac{4}{5}\)
\(\Rightarrow x=\dfrac{125}{16}\)
a, \(\dfrac{2}{3}\)\(x\) - \(\dfrac{1}{2}\)\(x\) = (- \(\dfrac{7}{12}\)). 1\(\dfrac{2}{5}\)
\(x\).(\(\dfrac{2}{3}\) - \(\dfrac{1}{2}\)) = (- \(\dfrac{7}{12}\)) . \(\dfrac{7}{5}\)
\(x\). \(\dfrac{1}{6}\) = - \(\dfrac{49}{60}\)
\(x\) = - \(\dfrac{49}{60}\).6
\(x\) = -\(\dfrac{49}{10}\)
2x + 3 + 2x = 144
<=> 2x (2^3 + 1) = 144
<=> 2x . 9 = 144
<=> 2x = 16
<=> x = 4
b/ x2 = 125 c/ x2 = 144
x2 = 52 x2 = 122
=> x = 5 hoặc x = -5 => x = 12 hoặc x = -12