1+2+3+4+...+x=91
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n: (-8)*x+17=-23
=>\(x\cdot\left(-8\right)=-23-17=-40\)
=>\(x\cdot8=40\)
=>\(x=\dfrac{40}{8}=5\)
o: \(\left(-15\right)\cdot x=10\left(-4\right)-5\)
=>\(x\cdot\left(-15\right)=-40-5=-45\)
=>\(x\cdot15=45\)
=>\(x=\dfrac{45}{15}=3\)
p: \(\left(-3\right)\cdot x-4=2\cdot\left(-7\right)+4\)
=>\(\left(-3\right)\cdot x-4=-14+4=-10\)
=>\(x\left(-3\right)=-10+4=-6\)
=>\(3x=6\)
=>\(x=\dfrac{6}{3}=2\)
q: x+x+x+91=-2
=>\(3x+91=-2\)
=>\(3x=-2-91=-93\)
=>\(x=-\dfrac{93}{3}=-31\)
r: \(-152-\left(3x+1\right)=\left(-2\right)\cdot\left(-27\right)\)
=>\(-152-3x-1=54\)
=>\(-153-3x=54\)
=>\(3x=-153-54=-207\)
=>\(x=-\dfrac{207}{3}=-69\)
5x + 2x = 91
7x = 91
x = 91 : 7 = 13
4x-3 = 256
4x-3 = 44
=> x - 3 = 4
=> x = 7
Hai câu sau không rõ, xin mời viết rõ ràng hơn
(1+1+1+.............+1+1):50+2010-12xX=91
(1x50):50+2010-12xX
50:50+2010-12xX=91
1+2010-12xX=91
2011-12xX=91
12xX=2011-91
12xX=1920
x=1920:12
x=160
( 100- 99+ 98- 97+ 96- 95+...+ 4- 3+ 2- 1): 50+ 2010- 12. X= 91.
A: 50+ 2010- 12. X= 91.
Xét A= 100- 99+ 98- 97+ 96- 95+...+ 4- 3+ 2- 1.
Dãy A có số các số hạng là:
( 100- 1): 1+ 1= 100( số)
Ta ghép 2 số hạng vào 1 nhóm được 50 nhóm.
A=( 100- 99)+( 98- 97)+( 96- 95)+...+( 4- 3)+( 2- 1).
A= 1+ 1+ 1+...+ 1+ 1( 50 số 1)
A= 1x 50.
A= 50.
=> A: 50+ 2010- 12. X= 91.
=> 50: 50+ 2010- 12. X= 91.
=> 1+ 2010- 12. X= 91.
=> 2011- 12. X= 91.
=> 12. X= 2011- 91.
=> 12. X= 1920.
=> X= 1920: 12.
=> X= 160.
Vậy X= 160.
Bài 4:
a: xy=-2
=>\(x\cdot y=1\cdot\left(-2\right)=\left(-2\right)\cdot1=\left(-1\right)\cdot2=2\cdot\left(-1\right)\)
=>\(\left(x,y\right)\in\left\{\left(1;-2\right);\left(-2;1\right);\left(-1;2\right);\left(2;-1\right)\right\}\)
b: \(\left(x-1\right)\left(y+2\right)=-3\)
=>\(\left(x-1\right)\cdot\left(y+2\right)=1\cdot\left(-3\right)=\left(-3\right)\cdot1=-1\cdot3=3\cdot\left(-1\right)\)
=>\(\left(x-1;y+2\right)\in\left\{\left(1;-3\right);\left(-3;1\right);\left(-1;3\right);\left(3;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(2;-5\right);\left(-2;-1\right);\left(0;1\right);\left(4;-3\right)\right\}\)
Bài 3:
a: \(x\left(x+9\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\x+9=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=0\\x=-9\end{matrix}\right.\)
b: \(\left(x-5\right)^2=9\)
=>\(\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=3+5=8\\x=-3+5=2\end{matrix}\right.\)
c: \(\left(7-x\right)^2=-64\)
mà \(\left(7-x\right)^2>=0\forall x\)
nên \(x\in\varnothing\)
Bài 2:
a: \(\left(-31\right)\cdot x=-93\)
=>\(31\cdot x=93\)
=>\(x=\dfrac{93}{31}=3\)
b: \(\left(-4\right)\cdot x=-20\)
=>\(4\cdot x=20\)
=>\(x=\dfrac{20}{4}=5\)
c: \(5x+1=-4\)
=>\(5x=-4-1=-5\)
=>\(x=-\dfrac{5}{5}=-1\)
d: \(-12x+1=-4\)
=>\(-12x=-4-1=-5\)
=>\(12x=5\)
=>\(x=\dfrac{5}{12}\)
a/ \(x-\dfrac{3}{7}=\dfrac{2}{5}\cdot\dfrac{1}{4}\)
\(x-\dfrac{3}{7}=\dfrac{1}{10}\)
\(x=\dfrac{1}{10}+\dfrac{3}{7}=\dfrac{37}{70}\)
Vậy....
b/ \(x+\dfrac{4}{5}=-\dfrac{5}{12}\cdot\dfrac{3}{25}\)
\(x+\dfrac{4}{5}=-\dfrac{1}{20}\)
\(x=-\dfrac{1}{20}-\dfrac{4}{5}=-\dfrac{17}{20}\)
Vậy....
c/ \(\dfrac{x}{182}=-\dfrac{6}{12}\cdot\dfrac{35}{91}\)
\(\dfrac{x}{182}=-\dfrac{5}{26}\)
\(=>x\cdot26=-5\cdot182\)
\(26x=-910\)
\(x=-910:26=-35\)
Vậy....
a) Ta có: \(x-\dfrac{3}{7}=\dfrac{2}{5}\cdot\dfrac{1}{4}\)
\(\Leftrightarrow x-\dfrac{3}{7}=\dfrac{1}{10}\)
\(\Leftrightarrow x=\dfrac{1}{10}+\dfrac{3}{7}=\dfrac{7}{70}+\dfrac{30}{70}\)
hay \(x=\dfrac{37}{70}\)
Vậy: \(x=\dfrac{37}{70}\)
\(3^{2x+1}=243\)
\(\Leftrightarrow3^{2x+1}=3^5\)
\(\Leftrightarrow2x+1=5\)
\(\Leftrightarrow2x=4\)
\(\Leftrightarrow x=2\)
\(P=\dfrac{\left(1+2+3+...+100\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{9}\right)\left(63\cdot1,2-21\cdot3,6\right)}{1-2+3-4+5-6+...+99-100}\)
đề là vậy nhé mn
để ý chút thấy liền ah : 63.1,2-21.3,6=63.1,2-21.3.1,2= 63.1,2- 63.1,2=0
=============================
Ta có P = \(\dfrac{\left(1+2+3+...+100\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+5-...+99-100}\)= \(\dfrac{\left(1+2+3+...+100\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{9}\right)0}{1-2+3-4+5-...+99-100}\)= \(\dfrac{0}{1-2+3-4+5-6+...+99-100}=0\)
Bài 1
a) -(515-80+91)-(2003+80-91)= -515+80-91-2003-80+91=(80-80)+(-91+91)+(-515-2003)=0+0+(-2518)= -2518
b)-(-85)-(-71)+15+(-85)=85+71+15+-85=(85-85)+(71+15)=0+86=86
c)-(537-812)+(-163-712)-(-364)= -537+812+-163-712+364=(812-712)+(-537-136)+364=0+(-700)+364= -236
Bài 2
a)25-(x+4)= -19+3 c) I2x-5l=13
25-(x+4)= -16 suy ra 2x-5 thuộc {13;-13}
(x+4)= 25-(-16) TH1: 2x-5=13 vậy x=9
x+4 =41 TH2: 2x-5= -13 vậy x= -4
x =41-4
x =37