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\(\frac{2}{1^2}.\frac{6}{2^2}.\frac{12}{3^2}.\frac{20}{4^2}...\frac{110}{10^2}.x=-20\)
\(\Leftrightarrow\frac{1.2}{1^2}.\frac{2.3}{2^2}.\frac{3.4}{3^2}.\frac{4.5}{4^2}...\frac{10.11}{10^2}.x=-20\)
\(\Leftrightarrow\frac{2}{1}.\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{11}{10}.x=-20\)
\(\Leftrightarrow11.x=-20\)
\(\Leftrightarrow x=\frac{-20}{11}\)
\(\dfrac{2}{1^2}.\dfrac{6}{2^2}.\dfrac{12}{3^2}.\dfrac{20}{4^2}.....\dfrac{110}{10^2}.x=-20\)
\(\dfrac{1.2}{1^2}.\dfrac{2.3}{2^2}.\dfrac{3.4}{3^2}.\dfrac{4.5}{4^2}.....\dfrac{10.11}{10^2}.x=-20\)
\(\dfrac{1.2.2.3.3.4.4.5.5.....10.10.11}{1.1.2.2.3.3.4.4.5.5.....10.10}.x=-20\)
\(11.x=-20\)
\(x=-20:11=-\dfrac{20}{11}\)
\(\dfrac{2}{1^2}\cdot\dfrac{6}{2^2}\cdot\dfrac{12}{3^2}\cdot\dfrac{20}{4^2}\cdot...\cdot\dfrac{110}{10^2}\cdot x=-20\)
\(\Leftrightarrow\dfrac{1\cdot2}{1\cdot1}\cdot\dfrac{2\cdot3}{2\cdot2}\cdot\dfrac{3\cdot4}{3\cdot3}\cdot\dfrac{4\cdot5}{4\cdot4}\cdot...\cdot\dfrac{10\cdot11}{10\cdot10}\cdot x=-20\)
\(\Leftrightarrow\dfrac{1\cdot2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot5\cdot...\cdot10\cdot11}{1\cdot1\cdot2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot...\cdot10\cdot10}\cdot x=-20\)
\(\Rightarrow11\cdot x=-20\)
\(\Rightarrow x=\dfrac{-20}{11}\)
Vậy \(x=\dfrac{-20}{11}\).
1) x - 36 + 12 = - x+ 10
=> x + x = 10 + 24
=> 2x = 34
=> x = 34/2 = 17
2) (x + 15) - (11 - x) = (-2)2
=> x + 15 - 11 + x = 4
=> 2x = 4 - 4
=> 2x = 0
=> x = 0
3) 40 - 4x2 = (-6)2
=> 40 - 4x2 = 36
=> 4x2 = 40 - 36
=> 4x2 = 4
=> x2 = 1
=> x = \(\pm\)1
4) (-50) + 10x2 = (-25) x |-2|
=> -50 + 10x2 = -50
=> 10x2 = -50 + 50
=> 10x2 = 0
=> x2 = 0
=> x = 0
5) |x + 1| = 2020
=> \(\orbr{\begin{cases}x+1=2020\\x+1=-2020\end{cases}}\)
=> \(\orbr{\begin{cases}x=2019\\x=-2021\end{cases}}\)
6) (x + 1)5 + 8 = 0 (xem lại đề)
7) (-20) + x3 : 16 = -24
=> x3 : 16 = -24 + 20
=> x3 : 16 = -4
=> x3 = -4 . 16
=> x3 = -64 = (-4)3
=> x = -4
9) x14 = x17
=> x14 - x17 = 0
=> x14(1 - x3) = 0
=> \(\orbr{\begin{cases}x^{14}=0\\1-x^3=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
10) (-36) + (1 - x)2 = 0
=> (1 - x)2 = 36
=> (1 - x)2 = 62
=> \(\orbr{\begin{cases}1-x=6\\1-x=-6\end{cases}}\)
=> \(\orbr{\begin{cases}x=-5\\x=7\end{cases}}\)
Bài 1:
a)a4.a5:a3=a4+5-3
= a6
b)m6.m.m7:m3
=m6+1+7-3
=m11
c)100.10000000:1000
=100.(107:103)
=102.107-3
=102.104
=102+4
=106
d)132-122
=13.13-12.12
=169-144
=25
=52
a. a^4 .a^5 :a^3
=\(a^9:a^3\)
=a\(^6\)
b)m6.m.m7:m3
=m\(^{14}\):m\(^3\)
=m^11
c)100.10000000:1000
=10^2.10^7:10^3
=10^9:10^3
10^6
d)13^2-12^2
=13.13-12.12
=169-144
=25
=5^2
bài 2:
a. 3(24-x^2)-10=14
3.(24-x^2)=14+10
3.(24-x^2)=24
24-x^2=24:3
24-x^2=8
x^2=24-8
x^2=16
x^2=4^2
Vậy x=4
b. (x-2)^3+2^3=6^2-1
(x-2)^3+2^3=.6.6.-1
(x-2)^3+2^3=35
(x-2)^3+8=35
(x-2)^3=35-8
(x-2)^3=27
(x-2)^3=3^3
x-2=3
x=3+2
x=5
Vậy x=5
c. 2^x+2\(^{x+3}\)=72
2^x+2\(^{x+3}\)=2^6+2^3
Vay x=6 và x=0
chúc bn học giỏi nha Thủy Trần
1) \(\Leftrightarrow x+11-15+x+20=0\)
\(\Leftrightarrow2x+16=0\)
\(\Leftrightarrow x=-8\)
2) \(\Leftrightarrow2x-16+x-13=16\)
\(\Leftrightarrow3x-45=0\)
\(\Leftrightarrow x=15\)
Những câu dưới bạn làm tương tự như vậy nhé
1)(x+11)–(15–x) =–20
x+11 - 15 + x = -20
x + ( 11 -15 ) = -20
x + ( -4 ) = -20
x = -20 - ( -4 )
x = -16
CHỊU THÔI KHÓ QUÁ KO BÍT LÀM !!!