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1. a) 4.415.8.25.125
= (4.25). (8.125).415
= 100.1000.415
= 100000.415
= 41500000
b) 2.31.12+4.42.6+8.27.3
= (2.31.12)+(4.42.6)+(8.27.3)
= (2.12).31+(4.6).42+(8.3).27
= 24.31+24.42+24.27
= 24 (31+42+27)
= 24.100
= 2400
Mình làm 5 câu thôi nhé !:
1) \(7^x\cdot49=7^{90}\)
\(\Rightarrow7^x\cdot7^2=7^{90}\)
\(\Rightarrow7^{x+2}=7^{90}\)
\(\Rightarrow x=90-2\)
\(\Rightarrow x=88\)
2) \(2^x\cdot4=128\)
\(\Rightarrow2^x\cdot2^2=2^7\)
\(\Rightarrow2^{x+2}=2^7\)
\(\Rightarrow x=7-2\)
\(\Rightarrow x=5\)
3) \(3^x-1=2^4\cdot5\)
\(\Rightarrow3^x=80+1\)
\(\Rightarrow3^x=81\)
\(\Rightarrow3^x=3^4\)
\(\Rightarrow x=4\)
4) \(3^x+15=42\)
\(\Rightarrow3^x=42-15\)
\(\Rightarrow3^x=27\)
\(\Rightarrow3^x=3^3\)
\(\Rightarrow x=3\)
5) \(4\cdot2^x-3=125\)
\(\Rightarrow2^2\cdot2^x=128\)
\(\Rightarrow2^{x+2}=2^7\)
\(\Rightarrow x=7-2\)
\(\Rightarrow x=5\)
6)
\(5^x=5^{15}:5^3\\ \Leftrightarrow5^x=5^{15-3}\\ \Leftrightarrow5^x=5^{12}\\ \Leftrightarrow x=12\)
7)
\(4^x=4^{15}:16\\ \Leftrightarrow4^x=4^{15}:4^2\\ \Leftrightarrow4^x=4^{15-2}\\ 4^x=4^{13}\\ \Leftrightarrow x=13\)
8)
\(7^x=7^{20}:7^{10}\\ \Leftrightarrow7^x=7^{20-10}\\ \Leftrightarrow7^x=7^{10}\\ \Leftrightarrow x=10\)
9)
\(11^x=11^{11}:11\\ \Leftrightarrow11^x=11^{11-1}\\ \Leftrightarrow11^x=11^{10}\\ \Leftrightarrow x=10\)
10)
\(3^{15}.3^x=3^{30}\\ \Leftrightarrow3^x=3^{30}:3^{15}\\ 3^x=3^{30-15}\\ \Leftrightarrow3^x=3^{15}\\ \Leftrightarrow x=15\)
1. 3x - 17 = x+3
3x - x = 3 + 17
2x = 20
x = 20 :2
x = 10
2. |x-3| - 12 = |-5|
|x-3| -12 = 5
|x-3| = 5 + 12
|x-3| = 17
=> x-3 = +17 , -17
- x - 3 = 17
x = 17 + 3
x = 20
- x - 3 = -17
x = -17 + 3
x = -14
3 . 25 - ( x -5 ) = -415 -( 15 -415)
25 - x +5 = -415 - 15 +415
25 - x +5 = ( -415 + 415) -15
25 - x +5 = 0-5 = -5
25 - x = -5-5
25 - x = -10
x = 25 - -10
x = 35
4. ( x -3 ) . ( 2x +6 ) = 0
=> x-3 = 0 hoặc 2x +6 = 0
- x - 3 = 0
x = 0 + 3 = 3
- 2x +6 = 0
2x = 0-6=-6
x = -6 : 2
x = -3
\(1;3x-17=x+3\)
\(\Leftrightarrow3x-x=3+17\)
\(\Leftrightarrow2x=20\)
\(\Rightarrow x=10\)
\(2;\left|x-3\right|-12=\left|-5\right|\)
\(\Leftrightarrow\left|x-3\right|-12=5\)
\(\Leftrightarrow\left|x-3\right|=5+12\)
\(\Leftrightarrow\left|x-3\right|=17\)
\(\Rightarrow\orbr{\begin{cases}x-3=17\\3-x=17\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=17+3\\-x=17-3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=20\\x=-14\end{cases}}}\)
\(3;25-\left(x-5\right)=-415-\left(15-415\right)\)
\(\Leftrightarrow25-x+5=-415-15+415\)
\(\Leftrightarrow-x=-415-15+415-25-5\)
\(\Leftrightarrow-x=-45\)
\(\Rightarrow x=45\)
\(4;\left(x-3\right)\left(2x+6\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\2x+6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}}\)
a)-Ix-3I+16=4
-Ix-3I=4-16
-Ix-3I=-12
-IxI=-12+3
-IxI=-9
x=-9
b)3-(17-x)=289-(36+289)
3-(17-x)=289-325
3-(17-x)=-36
(17-x)=3-(-36)
(17-x)=39
x=17-39
x=-22
c)25-(x+5)=415-(15-415)
25-(x+5)=415-(-400)
25-(x+5)=815
(x+5)=25-815
(x+5)=-790
x=(-790)-5
x=-795
d)34+(21-x)=(3747-30)-3746
34+(21-x)=3717-3746
34+(21-x)=-29
31+x=21-(-29)
31+x=50
x=50-31
x=19
\(2^8\cdot x-415=7^4\)
\(\Leftrightarrow256x-415=2401\)
\(\Leftrightarrow256x=2401+415\)
\(\Leftrightarrow256x=2816\)
\(\Leftrightarrow x=2816:256\)
\(\Leftrightarrow x=11\)
Bài 1: Tính
a) Ta có: \(\left(-25\right)\cdot68+\left(-34\right)\cdot\left(-250\right)\)
\(=-25\cdot68+\left(-340\right)\cdot\left(-25\right)\)
\(=-25\cdot\left(68-340\right)\)
\(=-25\cdot\left(-272\right)\)
\(=6800\)
b) Ta có: \(1999+\left(-2000\right)+2001+\left(-2002\right)\)
\(=1999-2000+2001-2002\)
\(=-1-1=-2\)
c) Ta có: \(515+\left[72+\left(-515\right)+\left(-32\right)\right]\)
\(=515+72-515-32\)
\(=40\)
d) Ta có: \(\left(2736-75\right)-2736+175\)
\(=2736-75-2736+175\)
\(=100\)
e) Ta có: \(-2020-\left(157-2020\right)-\left(-257\right)\)
\(=-2020-157+2020+257\)
\(=100\)
Bài 2: Tìm x
a) Ta có: \(x-\left|-2\right|=\left|-18\right|\)
\(\Leftrightarrow x-2=18\)
hay x=20
Vậy: x=20
b) Ta có: \(2x-\left|+14\right|=\left|-14\right|\)
\(\Leftrightarrow2x-14=14\)
\(\Leftrightarrow2x=28\)
hay x=14
Vậy: x=14
c) Ta có: \(\left|x+4\right|+5=20-\left(-12-7\right)\)
\(\Leftrightarrow\left|x+4\right|+5=20+12+7\)
\(\Leftrightarrow\left|x+4\right|=39-5=34\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=34\\x+4=-34\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=30\\x=-38\end{matrix}\right.\)
Vậy: x∈{30;-38}
d) Ta có: \(15-\left|2-x\right|=\left(-2\right)^2\)
\(\Leftrightarrow15-\left|2-x\right|=4\)
\(\Leftrightarrow\left|2-x\right|=11\)
\(\Leftrightarrow\left[{}\begin{matrix}2-x=11\\2-x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=13\end{matrix}\right.\)
Vậy: x∈{-9;13}
e) Ta có: \(\left|15-x\right|+\left|-25\right|=\left|-55\right|\)
\(\Leftrightarrow\left|15-x\right|+25=55\)
\(\Leftrightarrow\left|15-x\right|=30\)
\(\Leftrightarrow\left[{}\begin{matrix}15-x=30\\15-x=-30\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-15\\x=45\end{matrix}\right.\)
Vậy: x∈{-15;45}
g) Ta có: \(\left|17-\left(-4\right)\right|+\left|-24-\left(-5\right)\right|=\left|-x+3\right|\)
\(\Leftrightarrow\left|17+4\right|+\left|-24+5\right|=\left|3-x\right|\)
\(\Leftrightarrow\left|3-x\right|=40\)
\(\Leftrightarrow\left[{}\begin{matrix}3-x=40\\3-x=-40\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-37\\x=43\end{matrix}\right.\)
Vậy: x∈{-37;43}
a, \(\Leftrightarrow\)\(2x=-32\)
\(\Leftrightarrow\)\(x=-16\)
b,\(\Leftrightarrow\)\(25-\left(5x-3\right)=64\)
\(\Leftrightarrow\)\(5x-3=-39\)
\(\Leftrightarrow\)\(5x=-36\)
\(\Leftrightarrow\)\(x=-7,2\)
c,\(\Leftrightarrow\)\(25-\left(x-5\right)=15\)
\(\Leftrightarrow\)\(x-5=10\)
\(\Leftrightarrow\)\(x=15\)
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