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=( 9 + 13 +....+85 ) - ( x + x +.....+ x) =925
=940 - x.20 = 925
x.20 = 940-925
x.20 = 15
x = 15:20
x = 0.75
Ta có:
( 9 - x ) + ( 13 - x ) + ( 17 - x ) +....+ ( 85 - x ) = 925
=> 9 - x + 13 - x + 17 - x + .... + 85 - x = 925
=> ( 9 + 13 + 17 + .... + 85 ) - ( x + x + x + .... + x ) = 925
=> 940 - 20x = 925
=> x = ( 940 - 925 ) : 20
=> x = 3/4
\(\left(9-x\right)+\left(13-x\right)+\left(17-x\right)+...+\left(85-x\right)=925\)
\(9+13+17+...+85-x-x-x-...-x=925\)
Vì cứ 1 số hạng lại có 1x
Do đó số số hạng từ 9 đén 85 là:
( 85 - 9 ) : 4 + 1 = 20 ( số hạng )
Tổng số hạng từ 9 đén 85 là:
( 85 + 9 ) . 20 : 2 = 940
Vậy có -20x. Do đó ta đc:
\(9+13+17+...+85-x-x-x-...-x=925\)
\(240-20x=925\)
\(x=-34,25\)
Ko bt có đúng ko sai thì mk xin lỗi nha
a) \(x-\left(-15\right)=-13-\left(-85-13\right)\)
\(\Leftrightarrow x+15=-13-\left(-98\right)\Leftrightarrow x+15=-13+98=85\)
\(\Leftrightarrow x=85-15=70\)
b) \(\left(-9-x\right)+\left(x-14\right)=17-\left(-8+x\right)\)
\(\Leftrightarrow-9-x+x-14=17+8-x\Leftrightarrow-23=25-x\)
\(\Leftrightarrow-23-25=-x\Leftrightarrow-x=-48\Leftrightarrow x=48\)
Tự kết luận
a) x−(−15)=−13−(−85−13)
=> x+15=85
=>x=70
b) (−9−x)+(x−14)=17−(−8+x)
=>-9-x+x-14=17+8-x
=>-23=25-x
=>x=48
Bài 1:tính
a)65.(-19)+19.(-35)
=65.(-19)+(-19).35
=(-19).(65+35)
=(-19).100
=-1900
b)85.(35-27)-35.(85-27)
=85.35-85.27-35.85+35.27
=(85.35-35.85)+(-85.27+35.27)
=27.(-85+35)
=27.(-50)
=1350
c)47.(45-15)-47.(45+15)
=47.[(45-15)-(45+15)]
=47.[30-60]
=47.(-30)
=-1410
Bài2: Tìm các số nguyên x biết
a)(-2).(x+6)+6.(x-10)=8
-2x-12+6x-60=8
4x-72=8
4x=72+8
4x=50
x=\(\frac{25}{2}\)
b)(-4).(2x+9)-(-8x+3)-(x+13)=0
-6x-36+8x-3-x-13=0
x-41=0
x=41
Bài 1: Tính
a) \(65.\left(-19\right)+19.\left(-35\right)\)
= \(-1235+-665\)
= \(-1900\)
b) \(85.\left(35-27\right)-35.\left(85-27\right)\)
= \(-1350\)
c) \(47.\left(45-15\right)-47.\left(45+15\right)\)
=\(-1410\)
Bài 2: Tìm x:
\(\left(-2\right).\left(x+6\right)+6.\left(x-10\right)=8\)
\(x=20\)
Tìm x biết:
a,x-5/7=1/9
b,2x/5=6/2x+1
c,11/8+13/6=85/x
d,2x-2/11=1.1/5
e,x/15=3/5+-2/3
f,x/182=-6/14.35/91
a, \(x\) - \(\dfrac{5}{7}\) = \(\dfrac{1}{9}\)
\(x\) = \(\dfrac{1}{9}\) + \(\dfrac{5}{7}\)
\(x\) = \(\dfrac{52}{63}\)
b, \(\dfrac{2x}{5}\) = \(\dfrac{6}{2x+1}\)
2\(x\).(2\(x\) + 1) = 30
4\(x^2\)+ 2\(x\) - 30 = 0
4\(x^2\) + 12\(x\) - 10\(x\) - 30 = 0
(4\(x^2\) + 12\(x\)) - (10\(x\) + 30) =0
4\(x\).(\(x\) + 3) - 10.(\(x\) +3) = 0
2 (\(x\) + 3).(2\(x\) - 5) = 0
\(\left[{}\begin{matrix}x+3=0\\2x-5=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-3\\x=\dfrac{5}{2}\end{matrix}\right.\)
Vậy \(x\) \(\in\) {-3; \(\dfrac{5}{2}\)}
a) \(39\times118+29\times82\)
\(=39\times2\times59+29\times2\times41\)
\(=2\times\left(39\times59+29\times41\right)\)
\(=2\times\left(2301+1189\right)\)
\(=2\times3490\)
\(=6980\)
b) \(37\times39+78\times14-13\times85-52\times55\)
\(=37\times3\times13+6\times13\times14+13\times85-4\times13\times55\)
\(=13\times\left(37\times3+6\times14+85-4\times55\right)\)
\(=13\times\left(111+84+85-220\right)\)
\(=13\times\left(280-220\right)\)
\(=13\times60\)
\(=780\)
c) \(368:16+752:16+480:16\)
\(=\left(368+752+480\right):16\)
\(=\left(1120+480\right):16\)
\(=1600:16\)
\(=100\)
d) \(5\times11\times18+9\times31\times10+4\times29\times45\)
\(=5\times11\times9\times2+9\times31\times10+2\times2\times29\times5\times9\)
\(=10\times9\times11+10\times9\times31+10\times29\times2\times9\)
\(=90\times11+90\times31+90\times29\times2\)
\(=90\times\left(11+31+29\times2\right)\)
\(=90\times\left(42+58\right)\)
\(=90\times100\)
\(=9000\)
Giải:
Đặt \(A=\dfrac{6}{4.7}+\dfrac{6}{7.10}+\dfrac{6}{10.13}+...+\dfrac{6}{82.85}\)
\(\dfrac{1}{2}A=\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+...+\dfrac{3}{82.85}\)
\(\dfrac{1}{2}A=\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{82}-\dfrac{1}{85}\)
\(\dfrac{1}{2}A=\dfrac{1}{4}-\dfrac{1}{85}\)
\(\dfrac{1}{2}A=\dfrac{81}{340}\)
\(\Rightarrow A=\dfrac{81}{340}:\dfrac{1}{2}\)
\(A=\dfrac{81}{170}\)
\(A=\frac{34}{7.13}+\frac{51}{13.22}+\frac{85}{22.37}+\frac{68}{37.49}\)
\(=\frac{17.2}{7.13}+\frac{17.3}{13.22}+\frac{17.5}{22.37}+\frac{17.4}{37.49}\)
\(=17\left(\frac{2}{7.13}+\frac{3}{13.22}+\frac{5}{22.37}+\frac{4}{37.49}\right)\)
\(=\frac{17}{3}\left(\frac{6}{7.13}+\frac{9}{13.22}+\frac{15}{22.37}+\frac{12}{37.49}\right)\)
\(=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{22}+...+\frac{1}{37}-\frac{1}{49}\right)\)
\(=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{49}\right)\)\(=\frac{17}{3}.\frac{6}{49}=\frac{17.2}{49}=\frac{34}{49}\)
Có : \(\frac{17}{24}=\frac{34}{48}\)
Vì 48 < 49 => \(\frac{34}{48}>\frac{34}{49}\). Hay \(\frac{17}{24}>A\)
(9-x)+(13-x)+....+(85-x)=925
(=) 9-x+13-x+.....+85-x=925
(=) (9+13+.....+85)-(x+x+....x)=925
(=)940-20x=925
(=)20x=15
(=)x=\(\frac{3}{4}\)
(9-x) + ( 13 - x) + ......+ ( 85 - x ) = 925
=( 9 + 13 +....+85 ) - ( x + x +.....+ x) =925
=940 - x.20 = 925
x.20 = 940-925
x.20 = 15
x = 15:20
x = 0.75
cho mk 1tk nha