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a.(2x +1). (2x+1)=1
Mà chỉ có 1.1=1
Vậy 2x + 1=1
2x=1-1
2x=0
Suy ra: x= 0
Hoàng Khánh Thi thiếu nha.
a) (2x+1)2 = \(\left(\pm1\right)^2\)
=> 2x + 1 = 1 hoặc 2x + 1 = -1
=> 2x = 0 hoặc 2x = -2
=> x = 0 hoặc x = -1.
a) 35x+8=33
<=> 5x+8=3
<=>5x=3-8=-5
<=>x=-1
b) 28-x=25
<=> 8-x=5
<=> x=8-5=3
\(3^{5x+8}=27\)
\(3^{5x+8}=3^3\)
\(5x+8=3\)
\(5x=-5\)
\(x=-1\)
\(2^{8-x}=32\)
\(2^{8-x}=2^5\)
\(8-x=5\)
\(x=3\)
=.= hok tốt!!
c,2x2+(−6)3:27=0c,2x2+(-6)3:27=0
⇒2x2+(−216):27=0⇒2x2+(-216):27=0
⇒2x2+(−8)=0⇒2x2+(-8)=0
⇒2x2=0−(−8)⇒2x2=0-(-8)
⇒2x2=8⇒2x2=8
⇒x2=8:2⇒x2=8:2
⇒x2=4⇒x2=4
⇒{x2=22x2=(−2)2⇒{x2=22x2=(-2)2
⇒{x=2x=−2⇒{x=2x=-2
Vậy x∈{(−2);2}
a) \(\left(2x+3\right)^2=\frac{9}{21}\)
<=> \(\orbr{\begin{cases}2x+3=\frac{3}{11}\\2x+3=\frac{-3}{11}\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-1\frac{4}{11}\\x=-1\frac{7}{11}\end{cases}}\)
Vậy...
1, -x3+3x2-3x+1
=1-3x.12+3.1.x2-x3
=(1-3x)3
bài này là hằng đẳng thức số 5: (a-b)3=a3-3a2b+3ab2-b2
3, ta có:
x3+8y3=x3+(2y)3=(x+2y)(x2-2xy+4y2
đây là hằng đẳng thức số 6
1.\(45^{10}.5^{30}=45^{10}.125^{10}=\left(45.125\right)^{10}=5625^{10}\)
2.a. \(\left(2x-1\right)^3=-8\Leftrightarrow\left(2x-1\right)^3=\left(-2\right)^3\)
\(\Leftrightarrow2x-1=-2\Leftrightarrow x=-\frac{1}{2}\)
b.\(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{cases}}\)
c. \(\left(2x+3\right)^2=\frac{9}{121}\Leftrightarrow\orbr{\begin{cases}2x+3=\frac{3}{11}\\2x+3=-\frac{3}{11}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{15}{11}\\x=-\frac{18}{11}\end{cases}}\)
d.\(\left(3x-1\right)^3=-\frac{8}{27}=\left(-\frac{2}{3}\right)^3\)
\(\Leftrightarrow3x-1=-\frac{2}{3}\Leftrightarrow x=\frac{1}{9}\)
4.
a.\(99^{20}=\left(99^2\right)^{10}=9801^{10}\)
Do \(9801^{10}< 9999^{10}\Rightarrow99^{20}< 9999^{10}\)
b.\(3^{4000}=\left(3^2\right)^{2000}=9^{2000}\)
\(\Rightarrow3^{4000}=9^{2000}\)
c.\(2^{332}=\left(2^3\right)^{110}.2^2=8^{110}.4\)
\(3^{223}=\left(3^2\right)^{110}.3^3=\left(3^2\right)^{110}.9=9^{110}.9\)
Ta thấy \(4.8^{110}< 9.9^{110}\)
Vậy \(2^{332}< 3^{223}\)
a)\(27^6:9^3=\left(3^3\right)^6:\left(3^2\right)^3=3^{18}:3^6=3^{12}\)
b)\(24^n:2^{2n}=\left(2^3.3\right)^n:2^{2n}=2^{3n}.3^n:2^{2n}=2^{3n-2n}.3^n=2^n.3^n=6^n\)
c)\(32^4:8^6=\left(2^5\right)^4:\left(2^3\right)^6=2^{20}:2^{18}=2^2\)
(2x2 - 3)3=\(\frac{27}{8}\)
<=> (2x2 - 3)3=\(\left(\frac{3}{2}\right)^3\)
<=> 2x2 - 3 = \(\frac{3}{2}\)
<=> 2x2 =\(\frac{3}{2}+3\)
<=>2x2=\(\frac{9}{2}\)
<=> x2 = \(\frac{9}{2}:2\)
<=> x2=\(\frac{9}{4}\)
\(\Rightarrow x\in\left\{-\frac{3}{2};\frac{3}{2}\right\}\)
vậy...
\(\left(2x^2-3\right)^3=\frac{27}{8}\)
\(\Leftrightarrow2x^2-3=\frac{3}{2}\)
\(\Leftrightarrow2x^2=\frac{9}{2}\)
\(\Leftrightarrow x^2=\frac{9}{4}\)
\(\Leftrightarrow x=\pm\frac{3}{2}\)