X : 15+18960=19134
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H
3
HD
5 tháng 4 2020
Hhigh
sửa đề : \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)
áp dụng t/c dãy t/s = nhau
\(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=\frac{58}{3+4+5}=\frac{58}{12}=\frac{29}{6}\)
\(\frac{x}{3}=\frac{29}{6}\Rightarrow x=\frac{29}{2}\)
\(\frac{y}{4}=\frac{29}{6}\Rightarrow y=\frac{58}{3}\)
\(\frac{z}{5}=\frac{29}{6}\Rightarrow z=\frac{145}{6}\)
vậy ...
NT
1
AH
Akai Haruma
Giáo viên
30 tháng 7 2023
Lời giải:
$(x-15)-x.13=0$
$x-15-x.13=0$
$(x-x.13)-15=0$
$x(1-13)-15=0$
$x.(-12)-15=0$
$x.(-12)=15$
$x=15:(-12)=\frac{-5}{4}$
20 tháng 8 2021
1) \(x^3-8x+7=\left(x-1\right)\left(x^2+x-7\right)\)
2) \(x^3+8x^2-9=\left(x-1\right)\left(x^2+9x+9\right)\)
3) \(3x^3-4x+1=\left(x-1\right)\left(3x^2+3x-1\right)\)
4) \(x^4-3x^2+3x-1=\left(x-1\right)\left(x^3+x^2-2x+1\right)\)
5) \(x^4-5x^2+4=\left(x-1\right)\left(x-2\right)\left(x+1\right)\left(x+2\right)\)
20 tháng 8 2021
1: Ta có: \(x^3-8x+7\)
\(=x^3-x-7x+7\)
\(=x\left(x-1\right)\left(x+1\right)-7\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x-7\right)\)
2: Ta có: \(x^3+8x^2-9\)
\(=x^3-x^2+9x^2-9\)
\(=x^2\left(x-1\right)+9\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left(x^2+9x+9\right)\)
3: Ta có: \(3x^3-4x+1\)
\(=3x^3-3x-x+1\)
\(=3x\left(x-1\right)\left(x+1\right)-\left(x-1\right)\)
\(=\left(x-1\right)\left(3x^2+3x-1\right)\)
4: Ta có: \(x^4-3x^2+3x-1\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)-3x\cdot\left(x-1\right)\)
\(=\left(x-1\right)\cdot\left(x^3+x+x^2+1-3x\right)\)
\(=\left(x-1\right)\left(x^3+x^2-2x+1\right)\)
NA
10 tháng 11 2021
Bad -> Worse than
Clever -> More clever than
Convenient -> More convenient than
Far -> Farther than hoặc Further than
Fresh -> Fresher than
Friendly -> Friendlire than
Generous -> More generous than
Good -> Better than
Happy -> Happier than
High -> Higher than
Little -> less than
Long -> Longer than
Modern -> The most modern
Old -> The oldest
Soon -> The soonest
Ugly -> The ugliest
Warm -> The warmest
NT
3
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
8 tháng 12 2023
câu a, \(\dfrac{x}{x+1}\); \(\dfrac{x^2}{1-x}\); \(\dfrac{1}{x^2-1}\) (đk \(x\)≠ -1; 1)
\(x^2\) - 1 = ( \(x\) - 1).(\(x\) + 1)
\(\dfrac{x}{x+1}\) = \(\dfrac{x.\left(x-1\right)}{\left(x+1\right).\left(x-1\right)}\);
\(\dfrac{x^2}{1-x}\) = \(\dfrac{-x^2}{x-1}\)= \(\dfrac{-x^2.\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(\dfrac{1}{x^2-1}\) = \(\dfrac{1}{\left(x-1\right)\left(x+1\right)}\)
b, \(\dfrac{10}{x+2}\); \(\dfrac{5}{2x-4}\); \(\dfrac{1}{6-3x}\) (đk \(x\) ≠ -2; 2)
2\(x-4\) = 2.(\(x\) - 2); 6 - 3\(x\) = - 3.(\(x\) - 2)
\(\dfrac{10}{x+2}\) = \(\dfrac{10.2.3\left(x-2\right)}{2.3\left(x+2\right)\left(x-2\right)}\) = \(\dfrac{60\left(x-2\right)}{6\left(x-2\right)\left(x+2\right)}\)
\(\dfrac{5}{2x-4}\) = \(\dfrac{5.3\left(x+2\right)}{2.3\left(x-2\right).\left(x+2\right)}\) = \(\dfrac{15.\left(x+2\right)}{6.\left(x-2\right)\left(x+2\right)}\)
\(\dfrac{1}{6-3x}\) = \(\dfrac{-1}{3.\left(x-2\right)}\) = \(\dfrac{-1.\left(x+2\right)}{3.2.\left(x-2\right)\left(x+2\right)}\) = \(\dfrac{-2.\left(x+2\right)}{6.\left(x-2\right).\left(x+2\right)}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
8 tháng 12 2023
c, \(\dfrac{x}{2x-4}\); \(\dfrac{1}{2x+4}\) và \(\dfrac{3}{4-x^2}\) đk \(x\) ≠ 2; -2
\(\dfrac{x}{2x-4}\) = \(\dfrac{x}{2.\left(x-2\right)}\) = \(\dfrac{x.\left(x+2\right)}{2.\left(x-2\right).\left(x+2\right)}\)
\(\dfrac{1}{2x+4}\) = \(\dfrac{1}{2.\left(x+2\right)}\) = \(\dfrac{\left(x-2\right)}{2.\left(x+2\right).\left(x-2\right)}\)
\(\dfrac{3}{4-x^2}\) = \(\dfrac{-3}{\left(x-2\right)\left(x+2\right)}\) = \(\dfrac{-6}{2.\left(x-2\right)\left(x+2\right)}\)
HT
1
26 tháng 8 2021
Bài 4 :
\(n_{H2}=\dfrac{V_{H2}}{22,4}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
Pt : \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2|\)
2 3 1 3
0,1 0,15 0,05 0,15
a) \(n_{Al}=\dfrac{0,15.2}{3}=0,1\left(mol\right)\)
⇒ \(m_{Al}=n_{Al}.M_{Al}\)
= 0,1 . 27
= 2,7 (g)
\(m_{Cu}=10-2,7=7,3\left(g\right)\)
0/0Al = \(\dfrac{m_{Al}.100}{m_{hh}}=\dfrac{2,7.100}{10}=27\)0/0
0/0Cu = \(\dfrac{m_{Cu}.100}{m_{hh}}=\dfrac{7,3.100}{10}=13\)0/0
b) \(n_{Al2\left(SO4\right)3}=\dfrac{0,15.1}{3}=0,05\left(mol\right)\)
⇒ \(m_{Al2\left(SO4\right)3}=n_{Al2\left(SO4\right)3.}M_{Al2\left(SO4\right)3}\)
= 0,05 . 342
= 17,1 (g)
\(n_{H2SO4}=\dfrac{0,1.3}{2}=0,15\left(mol\right)\)
⇒ \(m_{H2SO4}=n_{H2SO4}.M_{H2SO4}\)
= 0,15 .98
= 14,7 (g)
\(C_{H2SO4}=\dfrac{m_{ct}.100}{m_{dd}}\Rightarrow m_{dd}=\dfrac{m_{ct}.100}{C}=\)\(\dfrac{14,7.100}{15}=98\left(g\right)\)
mdung dịch sau phản ứng = (mAl + mCu) + mH2SO4 - mH2
= 10 + 98 - (0,15 . 2)
=107,7 (g)
\(C_{Al2\left(SO4\right)3}=\dfrac{m_{ct}.100}{m_{dd}}=\dfrac{17,1.100}{107,7}=15,88\)0/0
Chúc bạn học tốt
X : 15 = 19134 - 18960
X : 15 = 174
X = 174 x 15
X = 2610
Bài tìm X ạ