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a) x2 + x = 0
=> x( x+ 1 ) = 0
=> x = 0
hoặc x = -1
b) b, (x-1)x+2 = (x-1)x+4
=> x + 2 = x + 4
=> 0x = 2 ( ktm)
Vậy ko có giá trị x nào thoả mãn đk
d) Ta có: x-1/x+5 = 6/7
=>(x-1).7 = (x+5).6
=>7x-7 = 6x+ 30
=> 7x-6x = 7+30
=> x = 37
Vậy x = 37
e, x2/ 6= 24/25
=> x2 . 25 = 6 . 24
⇒x2.25=144⇒x2.25=144
⇒x2=144÷25⇒x2=144÷25
⇒x2=5,76=2,42=(−2,42)⇒x2=5,76=2,42=(−2,42)
⇒x∈{2,4;−2,4}⇒x∈{2,4;−2,4}
Vậy x∈{2,4;−2,4}
\(\left(\frac{1}{2}\right)^5\times x=\left(\frac{1}{2}\right)^7\)
\(x=\left(\frac{1}{2}\right)^7\div\left(\frac{1}{2}\right)^5\)
\(x=\left(\frac{1}{2}\right)^{7-5}=\left(\frac{1}{2}\right)^2=\frac{1}{4}\) .
\(\left(\frac{3}{7}\right)^2\times x=\left(\frac{9}{21}\right)^2\)
\(\left(\frac{3}{7}\right)^2\times x=\left(\frac{3}{7}\right)^4\)
\(x=\left(\frac{3}{7}\right)^4\div\left(\frac{3}{7}\right)^2\)
\(x=\left(\frac{3}{7}\right)^{4-2}=\left(\frac{3}{7}\right)^2=\frac{9}{49}\)
\(2^x=2\Rightarrow x=1\)
\(3^x=3^4\Rightarrow x=4\)
\(7^x=7^7\Rightarrow x=7\)
\(\left(-3\right)^x=\left(-3\right)^5\Rightarrow x=5\)
\(\left(-5\right)^x=\left(-5\right)^4\Rightarrow x=4\)
\(2^x=4\Leftrightarrow2^x=2^2\Rightarrow x=2\)
\(2^x=8\Leftrightarrow2^x=2^3\Rightarrow x=3\)
\(2^x=16\Leftrightarrow2^x=2^4\Rightarrow x=4\)
\(3^{x+1}=3^2\Leftrightarrow x+1=2\Leftrightarrow x=2-1\Rightarrow x=1\)
\(5^{x-1}=5\Leftrightarrow x-1=1\Leftrightarrow x=1+1\Rightarrow x=2\)
\(6^{x+4}=6^{10}\Leftrightarrow x+4=10\Leftrightarrow x=10-4\Rightarrow x=6\)
\(5^{2x-7}=5^{11}\Leftrightarrow2x-7=11\Leftrightarrow2x=11+7\Leftrightarrow2x=18\Leftrightarrow x=18\div2\Rightarrow x=9\)
\(\left(-2\right)^{4x+2}=64\)
\(2^{-4x+2}=2^6\Leftrightarrow-4x+2=6\Leftrightarrow-4x=6-2\Leftrightarrow-4x=4\Leftrightarrow x=4\div\left(-4\right)\Rightarrow x=-1\)
\(\left(\frac{1}{2}\right)^x=\left(\frac{1}{2}\right)^5\Rightarrow x=5\)
\(\left(\frac{5}{6}\right)^{2x}=\left(\frac{5}{6}\right)^5\Rightarrow2x=5\Rightarrow x=\frac{5}{2}\)
\(\left(\frac{3}{4}\right)^{2x-1}=\left(\frac{3}{4}\right)^{5x-4}\Rightarrow2x-1=5x-4\)
\(2x-5x=-4+1\)
\(-3x=-3\Rightarrow x=1\)
\(\left(\frac{-1}{10}\right)^x=\frac{1}{100}\)
\(\left(\frac{1}{10}\right)^{-x}=\left(\frac{1}{10}\right)^2\Rightarrow-x=2\Rightarrow x=-2\)
\(\left(\frac{-3}{2}\right)^x=\frac{9}{4}\)
\(\left(\frac{3}{2}\right)^{-x}=\left(\frac{3}{2}\right)^2\Rightarrow-x=2\Rightarrow x=-2\)
\(\left(\frac{-3}{5}\right)^{2x}=\frac{9}{25}\)
\(\left(\frac{3}{5}\right)^{-2x}=\left(\frac{3}{5}\right)^2\Rightarrow-2x=2\Rightarrow x=-1\)
\(\left(\frac{-2}{3}\right)^x=\frac{-8}{27}\)
\(\left(\frac{-2}{3}\right)^x=\left(\frac{-2}{3}\right)^3\Rightarrow x=3\).
hehe.
đánh tới què tay, hoa mắt lun r nekkk!!
a, (x - 2)2 = 1
(x - 2)2 = -12
=> x - 2 = -1
x = -1 + 2
x = -1
b, (2x - 1)3 = -27
(2x - 1)3 = -33
=> 2x - 1 = -3
2x = -3 + 1
2x = -2
x = -2 : 2
x = -1
a) (x-2)^2 = 1 = 1^2 = (-1)^2
=> x-2 = 1 => x = 3
x - 2 = -1 => x = 1
.KL:..
b) (2x-1)^3 = -27 = (-3)^3
=> 2x-1 = -3 => 2x = -2 => x = -1
c)16/2^n = 1
2^4 : 2^n = 1
24-n = 1 = 20
=> 4-n = 0 => n = 4
c) (x-1/2)^3 = 1/27 = 1/3^3
=>x-1/2 = 1/3
x = 5/6
d) (x+1/2)^2 = 4/25 = (2/5)^2 = (-2/5)^2
...
rùi bn tự lm như phần a nha
e) (x-1)x+2 = (x-1)x+6
=> (x-1)x+2 - (x-1)x+6 = 0
(x-1)x+2.[1-(x-1)4 ] = 0
=> (x-1)x+2 = 0 => x-1 = 0 => x = 1
1-(x-1)4 = 0 => (x-1)^4 = 1 => x -1 = 1 => x = 2
x -1 = -1 => x = 0
KL:...
f) (x-2)2 + (y-3)2 = 0
=> (x-2)^2 = 0 => x - 2=0 => x = 2
(y-3)^2=0 => y-3 = 0 => y =3
g) 5(x-2).(x+3) = 1 = 50
=> (x-2).(x+3) = 0
=> x-2 = 0 => x = 2
x+3 = 0 => x = -3
KL:...
Bài 1:
a)
\(\dfrac{4^2\cdot25^2+32\cdot125}{2^3\cdot5^2}\\ =\dfrac{\left(2^2\right)^2\cdot\left(5^2\right)^2+2^5\cdot5^3}{2^3\cdot5^2}\\ =\dfrac{2^{2\cdot2}\cdot5^{2\cdot2}+2^5\cdot5^3}{2^3\cdot5^2}\\ =\dfrac{2^4\cdot5^4+2^5\cdot5^3}{2^3\cdot5^2}\\ =\dfrac{2^4\cdot5^4}{2^3\cdot5^2}+\dfrac{2^5\cdot5^3}{2^3\cdot5^2}\\ =2\cdot5^2+2^2\cdot5\\ =2\cdot25+4\cdot5\\ =50+20\\ =70\)
c)
\(\dfrac{\left(1-\dfrac{4}{9}-2\right)\cdot16}{\left(2-3\right)^{-2}}+12\\ =\dfrac{\left(\dfrac{9}{9}-\dfrac{4}{9}-\dfrac{18}{9}\right)\cdot16}{\left(-1\right)^{-2}}+12\\ =\dfrac{\dfrac{-13}{9}\cdot16}{\dfrac{1}{\left(-1\right)^2}}+12\\ =\dfrac{\dfrac{-208}{9}}{1}+12\\ =\dfrac{-208}{9}+12\\ =\dfrac{-208}{9}+\dfrac{108}{9}\\ =\dfrac{100}{9}\)
Bài 2:
a)
\(\left(x+2\right)^2=36\\ \Rightarrow\left[{}\begin{matrix}x+2=6\\x+2=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-8\end{matrix}\right.\)
b)
\(\left(1,78^{2x-2}-1,78^x\right):1,78^x=0\\ \Leftrightarrow\dfrac{1,78^{2x-2}}{1,78^x}-\dfrac{1,78^x}{1,78^x}=0\\ \Leftrightarrow\dfrac{1,78^{2x-2}}{1,78^x}-1=0\\ \Leftrightarrow \dfrac{1,78^{2x-2}}{1,78^x}=1\\ \Leftrightarrow1,78^{2x-2}=1,78^x\\ \Leftrightarrow2x-2=x\\ \Leftrightarrow2x-x=2\\ \Leftrightarrow x=2\)
d) \(5^{\left(x-2\right)\left(x+3\right)}=1\)
\(\Rightarrow5^{\left(x-2\right)\left(x+3\right)}=5^0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Vậy \(x_1=-3;x_2=2\)
Bài 1:
a) -6x + 3(7 + 2x)
= -6x + 21 + 6x
= (-6x + 6x) + 21
= 21
b) 15y - 5(6x + 3y)
= 15y - 30 - 15y
= (15y - 15y) - 30
= -30
c) x(2x + 1) - x2(x + 2) + (x3 - x + 3)
= 2x2 + x - x3 - 2x2 + x3 - x + 3
= (2x2 - 2x2) + (x - x) + (-x3 + x3) + 3
= 3
d) x(5x - 4)3x2(x - 1) ??? :V
Bài 2:
a) 3x + 2(5 - x) = 0
<=> 3x + 10 - 2x = 0
<=> x + 10 = 0
<=> x = -10
=> x = -10
b) 3x2 - 3x(-2 + x) = 36
<=> 3x2 + 2x - 3x2 = 36
<=> 6x = 36
<=> x = 6
=> x = 5
c) 5x(12x + 7) - 3x(20x - 5) = -100
<=> 60x2 + 35x - 60x2 + 15x = -100
<=> 50x = -100
<=> x = -2
=> x = -2
Bài 1:
a) 2|x-1| = 24.64
=> 2|x-1|= 210
=> |x-1|=10
=> \(\left[{}\begin{matrix}x-1=10\\x-1=-10\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=11\\x=-9\end{matrix}\right.\)
Vậy...
b)(3x-1)4=16
=> (3x-1)4=24
=> 3x - 1=2
=> 3x = 3
=> x=1
Vậy...
c) (2x+1)4=(2x+1)6
=> (2x+1)4 - (2x+1)6=0
=> (2x+1)4.[1 - (2x+1)2 ] = 0
=> \(\left[{}\begin{matrix}\left(2x+1\right)^4=0\\1-\left(2x+1\right)^2=0\end{matrix}\right.\)
+) (2x+1)4=04
=> 2x+1=0
=> 2x = -1
=> x= \(\frac{-1}{2}\)
+) 1 - (2x+1)2=0
=> (2x+1)2 = 1
=> \(\left[{}\begin{matrix}2x+1=1\\2x+1=-1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Vậy...
d) x13=27.x10
=> x3=33
=> x=3
e)2x+2x+3=144
=> 2x(1+8)=144
=> 2x= 16 = 24
=> x=4
Bài 2:
a) Hình như đề bài là thế này:
CMR: 55-54+53 chia hết cho 7
Xét 55-54+53
=53(52-5+1)
=53. 21
Mà 21\(⋮\)7 => 53.21 chia hết cho 7 hay 55-54+53
Vậy...
b) Xét 76+75-74
= 74(72+7-1)
=74.55
Mà 55 \(⋮\)11 => 74.55 chia hết cho 11 hay 76+75-74 chia hết cho 7
Vậy...
bài 1:
\(\left(\dfrac{3}{7}+\dfrac{1}{2}\right)^2=\left(\dfrac{6}{14}+\dfrac{7}{14}\right)^2=\left(\dfrac{13}{14}\right)^2=\dfrac{169}{196}\)
\(\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2=\left(\dfrac{9}{12}-\dfrac{10}{12}\right)^2=\left(-\dfrac{1}{12}\right)^2=\dfrac{1}{144}\)
\(\dfrac{5^4\cdot20^4}{25^5\cdot45}=\dfrac{5^4\cdot5^4\cdot\left(2^2\right)^4}{\left(5^2\right)^5\cdot5\cdot3^2}=\dfrac{5^8\cdot2^8}{5^{12}\cdot3^2}=\dfrac{1}{5^4}\cdot\dfrac{2^8}{3^2}=\dfrac{256}{5625}\)
Bài 2:
a: \(\left(x-1\right)^3=27\)
=>\(\left(x-1\right)^3=3^3\)
=>x-1=3
=>x=3+1=4
b: \(x^2+x=0\)
=>x(x+1)=0
=>\(\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
c: \(\left(2x+1\right)^2=25\)
=>\(\left[{}\begin{matrix}2x+1=5\\2x+1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=4\\2x=-6\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
d: \(\left(2x-1\right)^3=64\)
=>\(\left(2x-1\right)^3=4^3\)
=>2x-1=4
=>2x=4+1=5
=>\(x=\dfrac{5}{2}\)
1.Tính:
(3/7 + 1/2)^2:
(3/4 - 5/6)^2:
54 x 204 / 255 x 45:
Kết quả:
a. (x - 1)^3 = 27
b. x^2 + x = 0
c. (2x + 1)^2 = 25
d. (2x - 1)^3 = 64
Kết quả:
a. x = 4 b. x = 0 hoặc x = -1 c. x = 2 hoặc x = -3 d. x = 5/2