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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, nếu x<3/2suy ra x-2<0 suy ra |x-2|=-(x-2)=2-x
(3-2x)>0 suy ra|3-2x|=3-2x
ta có: 2-x+3-2x=2x+1
5-3x=2x+1
5-1=2x+3x
6=6x nsuy ra x=6(loại vì ko thuộc khả năng xét)
nếu \(\frac{3}{2}\le x<2\)thì x-2<0 suy ra|x-2|=-(x-2)=2-x
2-2x<0 suy ra|3-2x|=-(3-2x)=2x-3
ta có:2-x+2x-3=2x+1
-1+x=2x+1
-1-1=2x-x
-2=x(loại vì ko thuộc khả năng xét)
nếu \(x\ge2\)thì x-2\(\ge\)0suy ra:|x-2|=x-2
3-2x<0 suy ra:|3-2x|=-(3-2x)=2x-3
ta có:x-2+2x-3=2x+1
3x-5=2x+1
3x-2x=5+1
x=6(chọn vì thuộc khả năng xét)
suy ra x=6
c)\(tacó:2x=3y\Rightarrow\frac{x}{3}=\frac{y}{2}\Rightarrow\frac{x}{15}=\frac{y}{10}\)
\(4y=5z\Rightarrow\frac{y}{5}=\frac{z}{4}\Rightarrow\frac{y}{10}=\frac{z}{8}\)
suy ra:\(\frac{x}{15}=\frac{y}{10}=\frac{z}{8}=k\Rightarrow x=15k;y=10k;z=8k\)
ta có: 4(15k)-3(10k)+5(8k)=7
60k-30k+40k=7
70k=7 suy ra k=1/10
ta có:x=1/10.15=3/2
y=1/10.10=1
a , x.(2x+7)=0
(=) x = 0
2x + 7 = 0
(=) x = 0
2x = -7
(=) x = 0
x = -7/2
\(a,\left(2x-1\right)^3=-8\)
\(\Rightarrow\left(2x-1\right)^3=\left(-2\right)^3\)
\(\Rightarrow2x-1=-2\)
\(\Rightarrow2x=-1\)
\(\Rightarrow x=-\dfrac{1}{2}\)
\(b,\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
\(\Rightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{1}{4}\right)^2\)
\(\Rightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\)
\(\Rightarrow x=-\dfrac{1}{4}\)
\(c,\left(2x+3\right)^2=\dfrac{9}{121}\)
\(\Rightarrow\left(2x+3\right)^2=\left(\dfrac{3}{11}\right)^2\)
\(\Rightarrow2x+3=\dfrac{3}{11}\)
\(\Rightarrow2x=-\dfrac{30}{11}\)
\(\Rightarrow x=-\dfrac{15}{11}\)
\(d,\left(2x-1\right)^3=-\dfrac{8}{27}\)
\(\Rightarrow\left(2x-1\right)^3=\left(-\dfrac{2}{3}\right)^3\)
\(\Rightarrow2x-1=-\dfrac{2}{3}\)
\(\Rightarrow2x=\dfrac{1}{3}\Rightarrow x=\dfrac{1}{6}\)
\(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{1}{4}\right)^2\Leftrightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\Leftrightarrow x=\dfrac{-1}{4}\)
\(\left(2x+3\right)^2=\dfrac{9}{121}\Leftrightarrow\left(2x+3\right)^2=\left(\dfrac{3}{11}\right)^2\Leftrightarrow2x+3=\dfrac{3}{11}\Leftrightarrow x=\dfrac{-15}{11}\)
\(\left(2x-1\right)^3=-8\Leftrightarrow\left(2x-1\right)^3=\left(-2\right)^3\Leftrightarrow2x-1=-2\Leftrightarrow2x=-1\Leftrightarrow x=\dfrac{-1}{2}\)
a.
$7x-2y=5x-3y$
$\Leftrightarrow 2x=-y$. Thay vào điều kiện số 2 ta có:
$-y+3y=20$
$2y=20$
$\Rightarrow y=10$.
$x=\frac{-y}{2}=\frac{-10}{2}=-5$
b.
$2x=3y\Rightarrow \frac{x}{3}=\frac{y}{2}$
$3y=4z-2y\Rightarrow 5y=4z\Rightarrow \frac{y}{4}=\frac{z}{5}$
$\Rightarrow \frac{x}{6}=\frac{y}{4}=\frac{z}{5}$
Áp dụng tính chất dãy tỉ số bằng nhau:
$\frac{x}{6}=\frac{y}{4}=\frac{z}{5}=\frac{x+y+z}{6+4+5}=\frac{45}{15}=3$
$\Rightarrow x=6.3=18; y=4.3=12; z=5.3=15$
`x(2x-4)-(x-2)(2x+3)=0`
`<=>2x(x-2)-(x-2)(2x+3)=0`
`<=>(x-2)(2x-2x-3)=0`
`<=>(x-2)*(-3)=0`
`<=>x-2=0`
`<=>x=2`
TH1: \(x\ge5\)
<=> \(\left\{{}\begin{matrix}\left|x-5\right|=x-5\\\left|2x-1\right|=2x-1\end{matrix}\right.\)
PT <=> \(x-5+2x-1=2x+3\)
<=> x = 9 (Tm)
TH2: \(\dfrac{1}{2}\le x< 5\)
<=> \(\left\{{}\begin{matrix}\left|x-5\right|=5-x\\\left|2x-1\right|=2x-1\end{matrix}\right.\)
PT <=> 5 - x + 2x -1 = 2x + 3
<=> x = 1(Tm)
TH3: \(x< \dfrac{1}{2}\)
<=> \(\left\{{}\begin{matrix}\left|x-5\right|=5-x\\\left|2x-1\right|=1-2x\end{matrix}\right.\)
PT <=> \(5-x+1-2x=2x+3\)
<=> \(5x=3< =>x=\dfrac{3}{5}\left(l\right)\)
KL: x \(\in\left\{1;9\right\}\)
332 - 2x = 913
=> 332 : 32x = (32)13
=> 332 : 32x = 326
=> 32x = 332 : 326
=> 32x = 36
=>2x = 6
=> x = 3
\(3^{32-2X}=9^{13}\)
(=) \(3^{32-2X}=3^{26}\)
=>32-2X=26
(=)2X=6
(=)X=3