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a: (x-3)2=49
=>x-3=7 hoặc x-3=-7
=>x=10 hoặc x=-4
b: \(\left(x^4\right)^2=\dfrac{x^{12}}{x^5}\)
\(\Leftrightarrow x^8-x^7=0\)
\(\Leftrightarrow x^7\left(x-1\right)=0\)
=>x=0 hoặc x=1
c: \(\Leftrightarrow x^{10}-25x^8=0\)
\(\Leftrightarrow x^8\left(x^2-25\right)=0\)
\(\Leftrightarrow x^8\left(x-5\right)\left(x+5\right)=0\)
hay \(x\in\left\{0;5;-5\right\}\)
a) 5x . ( 53) 2 = 625
5x . 56 = 625
mà 625 = 54
Suy ra : x + 6 = 4
x = 4 - 6
x = -2
b) (-3/4 )3x - 1 = 256/81
(-3/4 )3X - 1 = (-3/4)-4
SUY RA : 3X - 1 = -4
3X = -4 + 1 = -3
X = -3 : 3
X = -1
C ) (8x - 1 )2n+1 = 52n+1
SUY RA : 8X - 1 = 5
8X = 5 + 1
8 X = 6
X = 6 : 8
X = 3/4
d) (x - 2/9 )2 = 4/9
mà 4/9 = 2/32
SUY RA : x - 2/9 = 2/3
x = 2/3 + 2/9
x = 24/27
Câu e mình không bít làm bn chịu khó suy nghĩ nha !
1/vì (1,782x-2-1,78x):1,78x=0
nên 1,78x2-2-1,78x=0
=>1,782x-2=1,78x
=>2x-2=x
2x=x+2
=>x=2
2/vì cơ số bằng nhau nên ta có
x-2=1;-1;0
ta có: x-2=1 => x=3
x-2=-1 => x=1
x-2=0 => x=2
3/ta có
(x+2)3=33 =>x+2=3 =>x=1
mik mệt rồi bạn cứ gải tiếp đi
a: \(\Leftrightarrow4^x\left(\dfrac{3}{2}+\dfrac{5}{3}\cdot4^2\right)=4^8\left(\dfrac{3}{2}+\dfrac{5}{3}\cdot4^2\right)\)
=>4^x=4^8
=>x=8
b: \(\Leftrightarrow2^x\cdot\dfrac{1}{2}+2^x\cdot2=2^{10}\left(2^2+1\right)\)
=>2^x=2^11
=>x=11
c: =>1/6*6^x+6^x*36=6^15(1+6^3)
=>6^x=6*6^15
=>x=16
d: \(\Leftrightarrow8^x\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)=8^9\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)\)
=>x=9
a) x : \(\left(-\frac{1}{3}\right)^3=-\frac{1}{3}\)
\(x:\frac{-1}{27}=\frac{-1}{3}\)
\(x=\frac{-1}{3}.\frac{-1}{27}\)
\(x=\frac{1}{81}\)
Vậy \(x=\frac{1}{81}\)
a) \(x:\left(-\frac{1}{3}\right)^3=-\frac{1}{3}\)
\(\Leftrightarrow x=\left(-\frac{1}{3}\right)\cdot\left(-\frac{1}{3}\right)^3\)
\(\Leftrightarrow x=\left(-\frac{1}{3}\right)^4\)
\(\Leftrightarrow x=\frac{1}{81}\)
b)\(\left(\frac{4}{5}\right)^5\cdot x=\left(\frac{4}{5}\right)^7\)
\(\Leftrightarrow x=\left(\frac{4}{5}\right)^7:\left(\frac{4}{5}\right)^5=\left(\frac{4}{5}\right)^2=\frac{16}{25}\)
c)\(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(\Leftrightarrow x+\frac{1}{2}=\frac{1}{4}\)
\(\Leftrightarrow x=-\frac{1}{4}\)
d)\(\left(3x+1\right)^3=-27\)
\(\Leftrightarrow3x+1=-3\)
\(\Leftrightarrow3x=-4\)
\(\Leftrightarrow x=-\frac{4}{3}\)
Bài 1 :
a. \(\left|x-\frac{1}{3}\right|< \frac{5}{2}\)
TH1 : nếu \(\left|x-\frac{1}{3}\right|>0\)
\(x-\frac{1}{3}< \frac{5}{3}\)
\(x< 2\)
TH2 : nếu \(\left|x-\frac{1}{3}\right|< 0\)
\(\frac{1}{3}-x< \frac{5}{3}\)
\(x>-\frac{4}{3}\)
Bài 2 :
a. \(\left(x-2\right)^2=1\)
\(\left(x-2\right)^2-1=0\)
\(\left(x-2-1\right)\left(x-2+1\right)=0\)
\(\left(x-3\right)\left(x-1\right)=0\)
\(\left[\begin{array}{nghiempt}x-3=0\\x-1=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=3\\x=1\end{array}\right.\)
a)Viết dưới dạng phân số rồi sử dụng tích chéo ý
b)\(\frac{-1}{7}.2^3-2x:1\frac{4}{3}=-2^{x-1}\)
\(\Rightarrow\frac{-8}{7}-2x:\frac{7}{3}=-2^{x-1}\)
\(\Rightarrow\frac{-8}{7}-\frac{6x}{7}=-2^{x-1}\)
\(\Rightarrow\frac{-8-6x}{7}=\frac{2^{x-1}}{-1}\)
\(\Rightarrow-1\left(-8-6x\right)=7.2^{x-1}\)
\(\Rightarrow6x+8=7.2^{x-1}\)
.........
1.
\(-3x^5y^4+3x^2y^3-7x^2y^3+5x^5y^4\)
\(=(-3x^5y^4+5x^5y^4)+(3x^2y^3-7x^2y^3)\)
\(=2x^5y^4-4x^2y^3\)
2.
\(\frac{1}{2}x^4y-\frac{3}{2}x^3y^4+\frac{5}{3}x^4y-x^3y^4\)
\(=(\frac{1}{2}x^4y+\frac{5}{3}x^4y)-(\frac{3}{2}x^3y^4+x^3y^4)\)
\(=\frac{13}{6}x^4y-\frac{5}{2}x^3y^4\)
3.
\(5x-7xy^2+3x-\frac{1}{2}xy^2\)
\(=(5x+3x)-(7xy^2+\frac{1}{2}xy^2)\)
\(=8x-\frac{15}{2}xy^2\)
4.
\(\frac{-1}{5}x^4y^3+\frac{3}{4}x^2y-\frac{1}{2}x^2y+x^4y^3\)
\(=(\frac{-1}{5}x^4y^3+x^4y^3)+(\frac{3}{4}x^2y-\frac{1}{2}x^2y)\)
\(=\frac{4}{5}x^4y^3+\frac{1}{4}x^2y\)
5.
\(\frac{7}{4}x^5y^7-\frac{3}{2}x^2y^6+\frac{1}{5}x^5y^7+\frac{2}{3}x^2y^6\)
\(=(\frac{7}{4}x^5y^7+\frac{1}{5}x^5y^7)+(-\frac{3}{2}x^2y^6+\frac{2}{3}x^2y^6)\)
\(=\frac{39}{20}x^5y^7-\frac{5}{6}x^2y^6\)
6.
\(\frac{1}{3}x^2y^5(-\frac{3}{5}x^3y)+x^5y^6=(\frac{1}{3}.\frac{-3}{5})(x^2.x^3)(y^5.y)+x^5y^6\)
\(=\frac{-1}{5}x^5y^6+x^5y^6=\frac{4}{5}x^5y^6\)
a,\(|x+1|+13=26\)
\(\Rightarrow|x+1|=26-13\)
\(\Rightarrow|x+1|=13\)
TH1: \(x+1=13\)
\(x=13-1\)
\(x=12\)
TH2: \(x+1=-13\)
\(x=\left(-13\right)-1\)
\(x=-14\)
b, \(3^x.\left(3^2+1\right)=250\)
\(3^x.10=250\)
\(3^x=250:10\)
\(3^x=25\)
c,