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Bài toán sai
bài toán đúng là x+(x+1)+(x+2)+(x+3)+...+19+20=20
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
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
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
3.
\(x_1+x_2+x_3+...+x_{2000}+x_{2001}+x_{2002}=0\)
\(\Rightarrow\left(x_1+x_2+x_3\right)+\left(x_4+x_5+x_6\right)+...+\left(x_{1999}+x_{2000}+x_{2001}\right)+x_{2002}=0\)
\(\Rightarrow1+1+...+1+x_{2002}=0\)
\(\Rightarrow\frac{\left(2001-1\right):1+1}{3}+x_{2002}=0\)
\(\Rightarrow667+x_{2002}=0\)
\(\Rightarrow x_{2002}=-667\)
Bài 1 :
a) \(2^3.15-\left[115-\left(12-5\right)^2\right]\)
\(=8.15-\left[115-7^2\right]=120-\left[115-49\right]\)
\(=120-66=54\)\
b) \(132-\left[116-\left(132-128\right)^2\right]\)
\(=132-\left[116-4^2\right]=132-100=32\)
c) \(\left[545-\left(45+4.25\right)\right]:50-2000:250+2^{15}:2^{13}\)
\(=\left[545-145\right]:50-2000:250+2^2\)
\(=400:50-2000:250+4\)
\(=8-8+4=4\)
d) \(\left[1104-\left(25.8+40\right)\right]:9+3^{16}:3^{12}\)
\(=\left[1104-240\right]:9+3^4\)
\(=864:9+81=96+81=177\)
Bài 2 :
a) \(575-\left(6x+70\right)=445\)
\(\Rightarrow\left(6x+70\right)=130\Rightarrow6x=60\Rightarrow x=10\)
b) \(315+\left(125-x\right)=435\)
\(\Rightarrow\left(125-x\right)=120\Rightarrow x=5\)
c) \(\left(3-x\right).\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3-x=0\\x-3=0\end{cases}\Rightarrow x=3}\)
d) e) .........................
Bài 3 :.....
Baif 4 :
a) \(A=x+\left\{-\left(x-2\right)-\left[x+\left(3-x\right)-\left(x+3\right)\right]\right\}\)
\(A=x+\left\{-x+2-\left[x+3-x-x-3\right]\right\}\)
\(A=x+\left\{-x+2-\left(-x\right)\right\}\)
\(A=x+\left\{-x+2+x\right\}=x+2\)
\(B=x.\left\{-\left(x-2\right)-\left[x+\left(3-x\right)-\left(x+3\right)\right]\right\}\)
\(B=x.\left\{-x+2-\left[x+3-x-x-3\right]\right\}\)
\(B=x.\left\{-x+2\left[-x\right]\right\}=x.\left\{-x+2+x\right\}=2x\)
b) Để A+B= 0
\(\Leftrightarrow x+2+2x=0\)
\(\Leftrightarrow3x+2=0\Leftrightarrow3x=-2\Leftrightarrow x=\frac{-2}{3}\)
Vậy tại x = \(\frac{-2}{3}\) thì A + B = 0
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