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\(\left(2x-1\right)\left(3x+5\right)+\left(-6x^3+5x\right):x\)
\(=6x^2+10x-3x-5-6x^3+5x:x\)
\(=-6x^3+6x^2+12x-5:x\)
\(=-6x^2+6x+12-\dfrac{5}{x}=-6\left(x^2-x-12\right)-\dfrac{5}{x}\)
A = (2\(x\) - 1)(3\(x\) + 5) + (-6\(x\)3 + 5\(x\)): \(x\)
A = 6\(x^2\) + 10\(x\) - 3\(x\) - 5 + \(x\)(- 6\(x^2\) + 5): \(x\)
A = 6\(x^2\) + 7\(x\) - 5 - 6\(x^2\) + 5
A = (6\(x^2\) - 6\(x^2\)) + 7\(x\) - (5 - 5)
A = 7\(x\)
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a) x=-213:(1+2+3+4+...+100)<=>x=-213/100
b) x-x=-1/3-2/4 <=> 0= -5/6 (vô lý )
c) x=-0,8119408369
d) x= 0.0258907758
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1)Tính:
a)\(4^2\cdot2=\left(2^2\right)^2\cdot2=2^4\cdot2=2^5=32\)
b)\(36^2:6^2=\left(36:6\right)^2=6^2=48\)
c)\(\left(\frac{2}{5}\right)^{10}:\left(\frac{4}{25}\right)^2=\left(\frac{2}{5}\right)^{10}\cdot\left(\frac{25}{4}\right)^2=\)\(\left(1\right)^{10}\cdot\left(\frac{5}{2}\right)^2=1\cdot\frac{5^2}{2^2}=1\cdot\frac{25}{4}=\frac{25}{4}\)
a
\(4^2.2=16.2=32\)
b\(36^2:6^2=36.36:6.6=36.36:36=36\)
c
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a. (x - 2)2 = 1
<=> (x - 2)2 = 12 = (-1)2
<=> \(\begin{cases}x-2=1\\x-2=-1\end{cases}\Leftrightarrow\begin{cases}x=3\\x=1\end{cases}\)
Vậy x \(\in\){1; 3}.
b. (2x - 1)3 = -8
<=> (2x - 1)3 = (-2)3
<=> 2x - 1 = -2
<=> 2x = -2 + 1
<=> 2x = -1
<=> x = -1/2
Vậy x = -1/2.
c. (x + 1/2)2 = 1/16
<=> (x + 1/2)2 = (1/4)2 = (-1/4)2
<=> \(\begin{cases}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{cases}\Leftrightarrow\begin{cases}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{cases}\)
Vậy x \(\in\){-1/4; -3/4}.
d. (x - 2)3 = -27
<=> (x - 2)3 = (-3)3
<=> x - 2 = -3
<=> x = -3 + 2
<=> x = -1
Vậy x = -1.
a.\(\left(x-2\right)^2\)=1
<=> x-2=1 hoặc x-2=-1
<=> x= 3 hoặc x=1
b.\(\left(2x-1\right)^3\)=-8
\(\left(2x-1\right)^3\)=\(\left(-2\right)^3\)
2x-1=-2
2x=-1
x=-1/2
c.\(\left(x+\frac{1}{2}\right)^2\)=\(\frac{1}{16}\)
\(\left(x+\frac{1}{2}\right)^2\)=\(\left(\frac{1}{4}\right)^2\)hoặc \(\left(x+\frac{1}{2}\right)^2\)=\(\left(-\frac{1}{4}\right)^2\)
x+\(\frac{1}{2}\)=\(\frac{1}{4}\) hoặc x+\(\frac{1}{2}\)=-\(\frac{1}{4}\)
x=-\(\frac{1}{4}\)hoặc x=-\(\frac{3}{4}\)
d.\(\left(x-2\right)^3\)=-27
\(\left(x-2\right)^3\)=\(\left(-3\right)^3\)
x-2=-3
x=-1
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a) \(\left(x-\frac{1}{2}\right)^3=27\)
=> \(\left(x-\frac{1}{2}\right)^3=3^3\)
=> \(x-\frac{1}{2}=3\)
=> \(x=3+\frac{1}{2}\)
=> \(x=\frac{7}{2}\)
Vậy \(x=\frac{7}{2}.\)
b) \(\left(2x-1\right)^3=-27\)
=> \(\left(2x-1\right)^3=\left(-3\right)^3\)
=> \(2x-1=-3\)
=> \(2x=\left(-3\right)+1\)
=> \(2x=-2\)
=> \(x=\left(-2\right):2\)
=> \(x=-1\)
Vậy \(x=-1.\)
Chúc bạn học tốt!