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\(x+\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+14\right)=14\)
\(x+x+1+x+2+x+3+...+x+14=14\)
\(15x+\left(1+2+3+...+14\right)=14\)
\(15x+\frac{\left(14+1\right).14}{2}=14\)
\(15x+105=14\)
\(15x=-91\)
\(x=-\frac{91}{15}=-6,06\)
Ko chắc lắm
x+(x+1)+(x+2)+(x+3)+...+(x+13)+(x+14)=14
x+x+1+x+2+x+3+...+x+13+x+14=14
(x+x+x+x+...+x+x)+(1+2+3+...+14)=14
15x + 14 = 14
15x = 14 - 14
15x = 0
x = 0:15
x = 0
Vậy x=0
\(x+\left(x+1\right)+\left(x+2\right)+....+13+14=14\Leftrightarrow x+\left(x+1\right)+....+13\Leftrightarrow x=-13\)
\(25+24+23+....+x+\left(x-1\right)+\left(x-2\right)=25\Leftrightarrow24+....+\left(x-2\right)=0\Leftrightarrow x-2=-24\)
\(\Leftrightarrow x=-22\)
Đáp án D
x + 1 4 - 1 3 : 2 + 1 6 - 1 4 = 7 46 = x + 1 4 - 1 3 : 23 12 = 7 46 x + 1 4 - 1 3 = 7 46 . 12 23 x + 1 4 - 1 3 = 7 24 x = 7 24 - 1 4 + 1 3 x = 3 8
Đáp án D
x + 1 4 - 1 3 : 2 + 1 6 - 1 4 = 7 46 = x + 1 4 - 1 3 : 23 12 = 7 46 x + 1 4 - 1 3 = 7 46 . 12 23 x + 1 4 - 1 3 = 7 24 x = 7 24 - 1 4 + 1 3 x = 3 8
Bài 1:
ta có 1314<1315
1313<1314
=> dấu cân điền là"<"(1314 -1313 <1315-1314)
a) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Vì \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne=\)
Nên x + 1 = 0 => x = -1
b) \(\frac{x+1}{14}+\frac{x+2}{13}=\frac{x+3}{12}+\frac{x+4}{11}\)
\(\Leftrightarrow\frac{x+1}{14}+1+\frac{x+2}{13}+1=\frac{x+3}{12}+1+\frac{x+4}{11}+1\)
\(\Leftrightarrow\frac{x+15}{14}+\frac{x+15}{13}=\frac{x+15}{12}+\frac{x+15}{11}\)
\(\Leftrightarrow\frac{x+15}{14}+\frac{x+15}{13}-\frac{x+15}{12}-\frac{x+15}{11}=0\)
\(\Leftrightarrow\left(x+15\right)\left(\frac{1}{14}+\frac{1}{13}-\frac{1}{12}-\frac{1}{11}\right)=0\)
Vì \(\frac{1}{14}+\frac{1}{13}-\frac{1}{12}-\frac{1}{11}\ne0\)
Nên x +15 = 0 => x = -15
a,\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Rightarrow\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)=\left(x+1\right).\left(\frac{1}{13}+\frac{1}{14}\right)\)
\(\Rightarrow\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)-\left(x+1\right).\left(\frac{1}{13}+\frac{1}{14}\right)=0\)
\(\Rightarrow\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Vì \(\frac{1}{10}>\frac{1}{13};\frac{1}{11}>\frac{1}{14}\Rightarrow\frac{1}{10}+\frac{1}{11}>\frac{1}{13}+\frac{1}{14}\Rightarrow\frac{1}{10}+\frac{1}{11}+\frac{1}{12}>\frac{1}{13}+\frac{1}{14}\)
\(\Rightarrow\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}>0\)
\(\Rightarrow x+1=0\Rightarrow x=-1\)
b, Bạn cộng thêm 1 vào \(\frac{x+1}{14};\frac{x+1}{13};\frac{x+1}{12};\frac{x+1}{11}\)Mội bên phân số 1 đơn vị rồi áp dụng như bài 1