Tìm x :
C ). ( 3 + 6 + 9 + 12 + 15 + 18 + 21 ) - x = 13
3 BN ĐẦU TIÊN ĐÚNG MK TIC NHA
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Bài 1:
a) 43+45+47+...+565 [có (565-43)/2+1= 262 số hạng)
= [(565+43)*262]/2
= 79648
b) 21+24+27+...+318 [có (318-21)/3+1= 100 số hạng]
= [(318+21)*100]/2
= 16950
tách ra từng cặp 1: 12/16 = x/4 => x= 12*4/16=3
12/16 = 21/y => y= 16*21/12=28
12/16 = z/80 => z= 12*80/16= 60
a)tách ra từng cặp 1: 12/16 = x/4 => x= 12*4/16=3
12/16 = 21/y => y= 16*21/12=28
12/16 = z/80 => z= 12*80/16= 60
a) \(3^x+6=9\Rightarrow3^x=3\Rightarrow3^x=3^1\Rightarrow x=1\)
b) \(x-3=\frac{-1}{2}\Rightarrow x=\frac{-1}{2}+3\Rightarrow x=\frac{5}{2}\)
c) \(\left|3-x\right|=\frac{1}{2}\Rightarrow\orbr{\begin{cases}3-x=\frac{1}{2}\\3-x=\frac{-1}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=\frac{7}{2}\end{cases}}\)
d) \(\left|x-2\right|+1=\frac{6}{4}\Rightarrow\left|x-2\right|=\frac{3}{2}-1\Rightarrow\left|x-2\right|=\frac{1}{2}\Rightarrow\orbr{\begin{cases}x-2=\frac{1}{2}\\x-2=\frac{-1}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=\frac{3}{2}\end{cases}}\)
a.x=1
b.x=5/2
c.x=35/12;x=37/12
d.x=5;x=3/2
Chúc bạn học tốt!
1) 5.( x - 6 ) - 2.( x + 9 ) = 21
5x - 30 - 2x - 18 = 21
3x - 48 = 21
3x = 21 + 48
3x = 69
x = 23
2) 2.( x + 3 ) + 3.( x + 1 ) = 15 - ( - 9 )
2x + 6 + 3x + 3 = 24
5x + 9 = 24
5x = 24 - 9
5x = 15
x = 3
3) ( - x + 5 ).(3 - x ) = 0
=> - x + 5 = 0 hoặc 3 - x = 0
=> x = 5 hoặc x = 3
4) ( x - 12 ) - 15 = ( 20 - 7 ) - ( 18 + x )
x - 12 - 15 = 13 - 18 - x
x - 27 = - 5 - x
x + x = - 5 + 27
2x = 22
x = 11
5) x - ( 17 - 8 ) = 5 + ( 10 - 3x )
x - 9 = 5 + 10 - 3x
x + 3x = 15 + 9
4x = 24
x = 6
a) chứng tỏ : abcabc chia hết cho 11
Ta có 123123:11=11193
Vậy abcabc chia hết cho 11
b)\(\frac{9\cdot15\cdot21\cdot12\cdot20}{5\cdot6\cdot45\cdot18\cdot4}=\frac{9\cdot3\cdot5\cdot3\cdot7\cdot2\cdot2\cdot3\cdot2\cdot2\cdot5}{5\cdot2\cdot3\cdot5\cdot3\cdot3\cdot2\cdot3\cdot3\cdot2\cdot2}\)\(=\frac{7\cdot5}{3}=\frac{35}{3}\)
a) \(\dfrac{x}{3}=\dfrac{-10}{6}\)
\(x\times6=-10\times3\)
\(x\times6=-30\)
\(x=-5\)
b) \(\dfrac{-8}{x}=\dfrac{-9}{15}\)
\(x\times-9=15\times-8\)
\(x\times-9=-120\)
\(x=\dfrac{40}{3}\)
c) \(\dfrac{2,7}{0,9}=\dfrac{-8}{x}\)
\(x\times2,7=-8\times0,9\)
\(x\times2,7=-7,2\)
\(x=-\dfrac{8}{3}\)
d) \(\dfrac{4}{9}=\dfrac{x}{12}\)
\(x\times9=12\times4\)
\(x\times9=48\)
\(x=\dfrac{48}{9}\)
\(x=\dfrac{16}{3}\)
Bài 46:
11: Ta có: \(-4\left|x-2\right|=-8\)
\(\Leftrightarrow\left|x-2\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=2\\x-2=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\)
Vậy: x∈{0;4}
12: Ta có: \(5\left|x+2\right|=-10\cdot\left(-2\right)\)
\(\Leftrightarrow5\left|x+2\right|=20\)
\(\Leftrightarrow\left|x+2\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)
Vậy: x∈{-6;2}
13: Ta có: \(6\left|x-2\right|=18:\left(-3\right)\)
\(\Leftrightarrow6\left|x-2\right|=-6\)(1)
Ta có: \(\left|x-2\right|\ge0\forall x\)
\(\Rightarrow6\left|x-2\right|\ge0\forall x\)(2)
Ta có: -6<0(3)
Từ (1), (2) và (3) suy ra x∈∅
Vậy: x∈∅
14: Ta có:\(-7\left|x+4\right|=21:\left(-3\right)\)
\(\Leftrightarrow-7\left|x+4\right|=-7\)
\(\Leftrightarrow\left|x+4\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=1\\x+4=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)
Vậy: x∈{-5;-3}
15: Ta có: \(4\left|x+1\right|=8\left(-2\right)-8\left(-5\right)\)
\(\Leftrightarrow4\left|x+1\right|=-16-\left(-40\right)\)
\(\Leftrightarrow4\left|x+1\right|=24\)
\(\Leftrightarrow\left|x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=6\\x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-7\end{matrix}\right.\)
Vậy: x∈{-7;5}
16: Ta có: \(3\left|x+5\right|=-9\)(4)
Ta có: |x+5|≥0∀x
⇒3|x+5|≥0∀x(5)
Ta có: -9<0(6)
Từ (4), (5) và (6) suy ra x∈∅
Vậy: x∈∅
17: Ta có: \(-8\left|x-3\right|=24-16:2\)
\(\Leftrightarrow-8\left|x-3\right|=16\)
\(\Leftrightarrow\left|x-3\right|=-2\)
mà |x-3|≥0>-2∀x
nên x∈∅
Vậy: x∈∅
18: Ta có: \(-3\left|x+6\right|=6\cdot2-9\)
\(\Leftrightarrow-3\left|x+6\right|=3\)
\(\Leftrightarrow\left|x+6\right|=-1\)
mà |x+6|≥0>-1∀x
nên x∈∅
Vậy: x∈∅
19: Ta có: \(5-\left|x+7\right|=4\)
\(\Leftrightarrow\left|x+7\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x+7=-1\\x+7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=-6\end{matrix}\right.\)
Vậy: x∈{-8;-6}
20: Ta có: \(12-\left|x+8\right|=10\)
\(\Leftrightarrow\left|x+8\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+8=2\\x+8=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=-10\end{matrix}\right.\)
Vậy: x∈{-10;-6}
câu a bằng 15/8
câu b bằng 19/6
câu c bằng 3/10
câu d bằng 27/4
(3+6+9+12+15+18+21)-x=13
84-x=13
x=71
#H
( 3 + 6 + 9 + 12 + 15 + 18 + 21 ) - x = 13
84-x =13
x =84-13
x =74
k nha