3a.(x-5)=59
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`@` `\text {Ans}`
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\(\dfrac{3}{5}\times\dfrac{96}{59}+\dfrac{3}{5}\times\dfrac{22}{59}\)
\(=\dfrac{3}{5}\times\left(\dfrac{96}{59}+\dfrac{22}{59}\right)\)
\(=\dfrac{3}{5}\times2\)
\(=\dfrac{6}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
15 × (4 - 7) - 15 × (5 - 3)
= 15 × (-3) - 15 × 2
= 15 × (-3 - 2)
= -15 × 5
= -75
--------
73 × (8 - 59) - 59 × (8 - 73)
= 73 × 8 - 73 × 59 - 59 × 8 + 59 × 73
= 8 × (73 - 59) + 73 × (59 - 59)
= 8 × 14 + 73 × 0
= 8 × 14
= 112
15 x (4-7) - 15 x (5-3)
= 15 x (-3) - 15 x (-2)
= 15 x (-3 - (-2))
= 15 x (-1)
= -15
73 x (8-59) - 59 x (8-73)
= 73 x 8 - 59 x 73 - 59 x (-65)
= 73 x 8 - 59 x (73 - (-65))
= 73 x 8 - 59 x 8
= 8 x (73-59)
= 8 x 14
= 112
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 2:
a: x-158=32
=>x=158+32
=>x=190
b: \(x\cdot24=264\)
=>\(x=\dfrac{264}{24}\)
=>x=11
c: \(6x+9=3^7:3^4\)
=>\(6x+9=3^3\)
=>6x+9=27
=>6x=18
=>x=18/6=3
Câu 1:
a: \(86\cdot19+14\cdot19\)
\(=19\left(86+14\right)\)
\(=19\cdot100=1900\)
b: \(4\cdot\left(-5\right)^2-104\cdot\left(-5\right)^2\)
\(=4\cdot25-104\cdot25\)
\(=25\left(4-104\right)=-100\cdot25=-2500\)
c: \(7\cdot\left(-2\right)\cdot8\left(-5\right)\)
\(=7\cdot2\cdot8\cdot5\)
\(=56\cdot10=560\)
d: \(59-\left[59+\left(-76\right)\right]\)
\(=59-59+76\)
=76
![](https://rs.olm.vn/images/avt/0.png?1311)
Giải:
(x+1)+(x+3)+...+(x+59)=1710
=>(x+x+x+...+x)+(1+3+..+59)=1710
=>30\(\times\)x+900=1710 (30 số x)
=>30\(\times\)x=810
=>x=810:30=27
Vậy x=27
Chúc bạn hok tốt
Bỏ phép tính sang 1 bên, ta lập dãy số:
1; 3; 5;...; 59
Quy luật: Mỗi số hạng liên tiếp sẽ cách nhau 2 đơn vị
⇒ Ta có số số hạng của dãy số cũng như số chữ số x là:
(59 - 1) : 2 + 1 = 30 (số)
Tổng của dãy là: (1 + 59) x 30 : 2 = 900
⇒ Ta lập lại biểu thức trên như sau:
X x 30 + 900 = 1710
X x 30 = 1710 - 900
X x 30 = 810
X = 27
Vậy số cần tìm là: 27
Lần sau ghi cả câu hỏi nhe, HT
![](https://rs.olm.vn/images/avt/0.png?1311)
( 1 + x ) + ( 3 + x ) + ( 5 + x ) + ... + ( 59 + x ) = 2010
( 1 + 3 + 5 + ... + 59 ) + ( x + x + x + ... + x ) = 2010
30x = 2010 - 900
30x = 1110
x = 1110 : 30
x = 37
<=>25x+1+3+5+7+...+59=2010
<=>25x+625 =2010
<=>25x =2010-625
<=>25x =1385
<=>x =1385:25
<=>x =55,4
Vậy x=55,4
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{x+1}{65}+\dfrac{x+3}{63}\) = \(\dfrac{x+5}{61}\) + \(\dfrac{x+7}{59}\)
<=> \(\dfrac{x+1}{65}+1+\dfrac{x+3}{63}+1\) = \(\dfrac{x+5}{61}\) + 1 + \(\dfrac{x+7}{59}\) + 1
<=> \(\dfrac{x+66}{65}+\dfrac{x+66}{63}\) = \(\dfrac{x+66}{61}\) + \(\dfrac{x+66}{59}\)
<=> \(\dfrac{x+66}{65}+\dfrac{x+66}{63}\) - \(\dfrac{x+66}{61}\) - \(\dfrac{x+66}{59}\) = 0
<=> (x + 66) . (\(\dfrac{1}{65}+\dfrac{1}{63}+\dfrac{1}{61}+\dfrac{1}{59}\)) = 0
<=> x + 66 = 0
<=> x = -66
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{x+1}{65}+\dfrac{x+3}{63}+\dfrac{x+5}{61}+\dfrac{x+7}{59}\)
\(\Leftrightarrow\dfrac{x+1}{65}+\dfrac{x+3}{63}-\dfrac{x+5}{61}-\dfrac{x+7}{59}=0\)
\(\left(\dfrac{x+1}{65}+1\right)+\left(\dfrac{x+3}{63}+1\right)-\left(\dfrac{x+5}{61}+1\right)-\left(\dfrac{x+7}{59}+1\right)\)
\(\Leftrightarrow\dfrac{x+66}{65}+\dfrac{x+66}{63}+\dfrac{x+66}{61}+\dfrac{x+66}{59}=0\)
\(\Leftrightarrow\left(x+66\right).\left[\left(\dfrac{1}{65}+\dfrac{1}{63}\right)-\left(\dfrac{1}{61}+\dfrac{1}{59}\right)\right]\)\(=0\)
Do \(\dfrac{1}{65}< \dfrac{1}{63}< \dfrac{1}{61}< \dfrac{1}{59}\)
\(\Rightarrow\left(\dfrac{1}{65}+\dfrac{1}{63}\right)-\left(\dfrac{1}{61}+\dfrac{1}{59}\right)< 0\)
Vậy để \(\left(x+66\right).\left[\left(\dfrac{1}{65}+\dfrac{1}{63}\right)-\left(\dfrac{1}{61}+\dfrac{1}{59}\right)\right]=0\)
\(\Leftrightarrow x+66=0\)
\(\Leftrightarrow x=-66\)
Vậy \(x\in\left\{-66\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(20\%\times25,5+\dfrac{1}{5}\times15,5+0,2\times59\)
\(=\dfrac{1}{5}\times25,5+\dfrac{1}{5}\times15,5+\dfrac{1}{5}\times59\)
\(=\dfrac{1}{5}\times\left(25,5+15,5+59\right)\)
\(=\dfrac{1}{5}\times100\)
\(=20\)