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Câu 1:
a: =>37-39+x=13-13-17
=>x-2=-17
=>x=-15
b: |x-3|+x=3
=>|x-3|=3-x
=>x-3<=0
=>x<=3
c: (x2+7)(x2-49)=0
=>(x+7)(x-7)=0
=>x=-7 hoặc x=7
\(\frac{-5}{6}+\frac{8}{3}+\frac{29}{-6}\le x\le\frac{-1}{2}+2+\frac{5}{2}\)
\(\Rightarrow-3\le x\le4\)
\(\Rightarrow x\in\left\{-3;-2;-1;0;1;2;3;4\right\}\)
\(\left(\frac{-2}{3}-\frac{1}{2}\right):\frac{-1}{4}\le x\le\left(\frac{-5}{6}+\frac{9}{4}:\frac{-3}{2}\right)\cdot\frac{-13}{2}\)
\(\Rightarrow\frac{14}{3}\le x\le\frac{91}{6}\)
\(\Rightarrow\frac{28}{6}\le x\le\frac{91}{6}\)
\(\Rightarrow x\in\left\{\frac{28}{6};\frac{29}{6};...;\frac{90}{6};\frac{91}{6}\right\}\)
a) \(37-\left(39-x\right)=\left|-13\right|-\left(13+17\right)\)
\(\Leftrightarrow37-39+x=13-30\)
\(\Leftrightarrow-2+x=-17\)
\(\Leftrightarrow x=-17+2\)
\(\Leftrightarrow x=-15\)
b) \(\left|x-3\right|+x=3\)
\(\Leftrightarrow\left|x-3\right|=3-x\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=3-x\\x-3=-3+x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x+x=3+3\\x-x=-3+3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=6\\0=0\left(L\right)\end{cases}}\)
\(\Leftrightarrow x=\frac{6}{2}\)
\(\Leftrightarrow x=3\)
c) \(\left(x^2+7\right)\left(x^2-49\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+7=0\\x^2-49=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0-7\\x^2=0+49\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=-7\\x^2=49\end{cases}}\)
Vì \(x^2\ge0\)
Mà \(-7< 0\)
\(\Rightarrow x^2=49\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=7^2\\x^2=\left(-7\right)^2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=7\\x=-7\end{cases}}\)
Ta có : a1 + a2 + a3 + a4 = a5 + a6 + a7 + a8 = ... = a997 + a998 + a999 + a1000 = 1
<=> a1 + a2 + a3 + a4 + a5 + a6 + a7 + a8 + ... + a997 + a998 + a999 + a1000 = 1 x 250
<=> a1 + a2 + a3 + a4 + a5 + a6 + a7 + a8 + ... + a997 + a998 + a999 + a1000 = 250
Ta có :( a1 + a2 + ... + a1001 ) - ( a1 + a2 + ... + a1000 ) = 0 - 250
<=> a1001 = -250
a: \(S=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}=-\dfrac{1}{100}\)
c: \(5S_3=5^6+5^7+...+5^{101}\)
\(\Leftrightarrow4\cdot S_3=5^{101}-5^5\)
hay \(S_3=\dfrac{5^{101}-5^5}{4}\)
d: \(S_4=7\cdot\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+...+\dfrac{1}{69}-\dfrac{1}{70}\right)\)
\(=7\left(\dfrac{1}{10}-\dfrac{1}{70}\right)=7\cdot\dfrac{6}{70}=\dfrac{6}{10}=\dfrac{3}{5}\)
1. thực hiện phép tính
a, 23. 15 - [ 115 - ( 12-5)2 ]
= 23 . 15 - [ 115 - 72 ]
= 8 . 15 - 66
= 120 - 66
= 54
b,132 - [ 116 - (132 - 128)2
= 132 - [ 116 - 42 )
= 132 - 100
= 32
c, [ 545 - ( 45 + 4.25 ) ] : 50 - 2000: 250 +215: 213
= [ 545 - 145 ] : 50 -8 + 22
= 400 : 50 - 8 + 4
= 8 - 8 + 4
= 4
d, [ 1104 - ( 25.8 + 40)] :9 + 316: 312
= [ 1104 - { 200+40 } ] : 9 + 34
= { 1104 - 240 ) : 9 + 81
= 864 : 9 + 81
= 177
2.tìm x bt
a, 575 - ( 6x + 70) = 445
=> 6x +70 = 575 - 445
=> 6x + 70 = 130
=> 6x = 130 - 70
=> 6x = 60
=> x = 60:6
=> x = 10
Vậy x = 10
b, 315 + (125 - x) = 435
=> 125 - x = 435-315
=> 125-x = 120
=> x = 125-120
=> x = 5
Vậy x = 5
c, (3-x).(x-3)=0
=> \(\orbr{\begin{cases}3-x=0\\x-3=0\end{cases}}\)=> \(\orbr{\begin{cases}x=3-0\\x=0+3\end{cases}}\)=> \(\orbr{\begin{cases}x=3\\x=3\end{cases}}\)
Vậy x = 3
Câu 4:
a) Ta có: \(\left|-x+8\right|\ge0\)
\(\Rightarrow A=\left|-x+8\right|-21\ge-21\)
Vậy \(MIN_A=-21\) khi x = 8
b) Ta có: \(\left|-x-17\right|+\left|y-36\right|\ge0\)
\(\Rightarrow B=\left|-x-17\right|+\left|y-36\right|+12\ge12\)
Vậy \(MIN_B=12\) khi \(x=-17;y=36\)
c) Ta có: \(-\left|2x-8\right|\le0\)
\(\Rightarrow C=-\left|2x-8\right|-35\le-35\)
Vậy \(MAX_C=-35\) khi \(x=4\)
d) Ta có: \(3\left(3x-12\right)^2\ge0\)
\(\Rightarrow D=3\left(3x-12\right)^2-37\ge-37\)
Vậy \(MIN_D=-37\) khi x = 4
e) Ta có: \(-3\left|2x+50\right|\le0\)
\(\Rightarrow E=-21-3\left|2x+50\right|\le-21\)
Vậy \(MAX_E=-21\) khi x = -25
g) \(\left(x-3\right)^2+\left|x^2-9\right|\ge0\)
\(\Rightarrow G=\left(x-3\right)^2+\left|x^2-9\right|+25\ge25\)
Vậy \(MIN_G=25\) khi x = 3
\(a,x^2=4\Rightarrow x^2=2^2\Rightarrow x=2\)
\(b,x^2=64\Rightarrow x^2=8^2\Rightarrow x=8\)
\(c,6x^3-8=40\Rightarrow6x^3=48\Rightarrow x^3=8\Rightarrow x^3=2^3\Rightarrow x=2\)
\(d,\left(2x-1\right)^2=49\Rightarrow\left(2x-1\right)^2=7^2\Rightarrow2x-1=7\Rightarrow x=4\)
\(e,2^x:16=2^5\Rightarrow2^x:16=32\Rightarrow2^x=512\Rightarrow2^x=2^9\Rightarrow x=9\)
\(f,4^5:4^x=16\Rightarrow1024:4^x=16\Rightarrow4^x=64\Rightarrow4^x=4^3\Rightarrow x=3\)
a, x^2 = 4
=> x = 2 hoặc x = -2
b, x^2 = 64
=> x = 8 hoặc x = -8
c, 6x^3 - 8 = 40
=> 6x^3 = 48
=> x^3 = 8
=> x = 2
d, (2x - 1)^2 = 49
=> 2x - 1 = 7 hoặc 2x - 1 = -7
=> 2x = 8 hoặc 2x = -6
=> x = 4 hoặc x = -3
e, 2^x : 16 = 2^5
=> 2^x : 2^4 = 2^5
=> 2^x = 2^9
=> x = 9
f, 4^5 : 4^x = 16
=> 4^5 - x = 4^2
=> 5 - x = 2
=> x = 3
\(a,x+\frac{1}{2}=\frac{3}{4}.\frac{8}{6}\)
\(x+\frac{1}{2}=1\)
\(x=\frac{1}{2}\)
\(b,x-\left(\frac{1}{6}+\frac{1}{3}\right)=\frac{5}{2}.\frac{7}{5}\)
\(x-\frac{1}{6}-\frac{1}{3}=\frac{7}{2}\)
\(x-\frac{1}{6}=\frac{23}{6}\)
\(x=4\)
Câu c,d đâu