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Đặt \(A=\frac{1.2+2.4+3.6+4.8+5.10}{4+6.8+9.12+12.16+15.20}\)
\(\Rightarrow A=\frac{1.\left(1+2\right)+2.\left(1+2\right)+3.\left(1+2\right)+4.\left(1+2\right)+5.\left(1+2\right)}{4+2.\left(3+4\right)+3.\left(3+4\right)+4.\left(3+4\right)+5.\left(3+4\right)}\)
\(\Rightarrow A=\frac{\left(1+2+3+4+5\right).\left(1+2\right)}{4+\left(2+3+4+5\right).\left(3+4\right)}\)
\(\Rightarrow A=\frac{\left(1+5\right).5:2.\left(1+2\right)}{4+\left(2+5\right).4:2.\left(3+4\right)}\)
\(\Rightarrow A=\frac{6.5:2.3}{4+7.4:2.7}=\frac{45}{4+98}=\frac{45}{102}=\frac{15}{34}\)
Vậy \(A=\frac{15}{34}\)
Chúc bn học tốt
Rút gọn biểu thức 5^2*6^11*16^2+6^2*12^6*15^2 / 2*6^12*10^4-81^2*960^3
Làm ơn giúp tớ, tớ cần rất gấp
\(\left(\dfrac{6:\dfrac{3}{5}-\dfrac{17}{16}.\dfrac{6}{7}}{\dfrac{21}{5}.\dfrac{10}{11}+\dfrac{57}{11}}-\dfrac{\left(\dfrac{3}{20}+\dfrac{1}{2}-\dfrac{1}{15}\right).\dfrac{12}{49}}{\dfrac{10}{3}+\dfrac{2}{9}}\right).x=\dfrac{215}{96}\)
\(\Rightarrow\left(\dfrac{\dfrac{509}{56}}{9}-\dfrac{\dfrac{7}{12}.\dfrac{12}{49}}{\dfrac{32}{9}}\right).x=\dfrac{215}{96}\)
\(\Rightarrow\left(\dfrac{509}{504}-\dfrac{\dfrac{1}{7}}{\dfrac{32}{9}}\right).x=\dfrac{215}{96}\)
\(\Rightarrow\left(\dfrac{509}{504}-\dfrac{9}{224}\right).x=\dfrac{215}{96}\)
\(\Rightarrow\dfrac{1955}{2016}.x=\dfrac{215}{96}\)
\(\Rightarrow x=\dfrac{215}{96}:\dfrac{1955}{2016}\)
\(\Rightarrow x=\dfrac{903}{391}\)
`[ 6 : 3/5 - 17/16 . 6/7 : 21/5 . 10/11 + 57/11 - (3/20 + 1/2 - 1/15) . 12/49 : 10/3 + 2/9 ] . x = 215/96`
`=>[ 6 . 5/3 - 17/16 . 6/7 . 5/21 . 10/11 + 57/11 - (3/20 + 1/2 - 1/15) . 12/49 . 3/10 + 2/9 ] . x = 215/96`
`=>[10- 51/56 . 6/7 . 5/21 . 10/11 + 57/11 - (3/20 + 1/2 - 1/15) . 12/49 . 3/10 + 2/9 ] . x = 215/96`
`=> [10- 153/196 . 5/21 . 10/11 + 57/11 - (3/20 + 1/2 - 1/15) . 12/49 . 3/10 + 2/9 ] . x = 215/96`
`=> [10- 255/1372 . 10/11 + 57/11 - (3/20 + 1/2 - 1/15) . 12/49 . 3/10 + 2/9 ] . x = 215/96`
`=> [10- 1275/7546 + 57/11 - (3/20 + 1/2 - 1/15) . 12/49 . 3/10 + 2/9 ] . x = 215/96`
`=> (10- 1275/7546 + 57/11 - 7/12. 12/49 . 3/10 + 2/9 ) . x = 215/96`
`=> ( 10- 1275/7546 + 57/11 -343/600 . 3/10 + 2/9 ) . x = 215/96`
`=> ( 10- 1275/7546 + 57/11 -343/2000 + 2/9 ) . x = 215/96`
`=>15,06357671 . x= 215/96`
`=> x= 215/96: 15,06357671`
`=>x= 0,1486754027`
Đề có phải như vậy không nhỉ ?
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}