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Câu 1 : \(\frac{x+2}{18}+\frac{x+2}{19}+\frac{x+2}{20}=\frac{x+2}{21}+\frac{x+2}{22}\)
=> \(\frac{x+2}{18}+\frac{x+2}{19}+\frac{x+2}{20}-\frac{x+2}{21}-\frac{x+2}{22}=0\)
=> x+2 . ( \(\frac{1}{18}+\frac{1}{19}+\frac{1}{20}-\frac{1}{21}-\frac{1}{22}\)) = 0
Vì \(\frac{1}{18}+\frac{1}{19}_{ }+\frac{1}{20}-\frac{1}{21}-\frac{1}{22}\ne0\)nên x+2=0
=> x= 0 - 2 = -2
Vậy x = -2
X sẽ bằng 2007 vì:
2032-x/25+2053-x/21+2070-x/21+2038-x/19 = 10 ( vì đỏi vế số 10 nên = 0+10=10)
10= 1+2+3+4 (Có 4 phân số thì mỗi phân số tương ứng lần lượt la 1 ,2 ,3 ,4)
Vậy x =2007
Chúc bạn học giỏi
=>\(\left(\frac{2032-x}{25}-1\right)+\left(\frac{2053-x}{23}-2\right)+\left(\frac{2070-x}{21}-3\right)+\left(\frac{2083-x}{19}-4\right)=0\)
=>\(\frac{2007-x}{25}+\frac{2007-x}{23}+\frac{2007-x}{21}+\frac{2007-x}{19}=0\)
=>\(\left(2007-x\right)\left(\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\right)=0\)
Vì \(\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\ne0\)
=> 2007 - x = 0 => x = 2007
a) \(6.8^{x-1}+8^{x+1}=6.8^{19}+8^{21}\)
\(\Rightarrow x-1+x+1=19+21\)
\(=2x=40\)
\(\Rightarrow x=20\)
b) \(4.3^{x-1}+2.3^{x+2}=4.3^6+2.3^9\)
\(\Rightarrow x-1+x+2=6+9\)
\(\Rightarrow2x+1=15\)
\(\Rightarrow2x=14\)
\(\Rightarrow x=7\)
a: \(\dfrac{2032-x}{25}+\dfrac{2053-x}{23}+\dfrac{2070-x}{21}+\dfrac{2083-x}{19}-10=0\)
\(\Leftrightarrow\left(\dfrac{2032-x}{25}-1\right)+\left(\dfrac{2053-x}{23}-2\right)+\left(\dfrac{2070-x}{21}-3\right)+\left(\dfrac{2083-x}{19}-4\right)=0\)
=>2007-x=0
hay x=2007
b: \(\Leftrightarrow x+\left(1+1+1+1+1+1+1\right)+\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)=0\)
\(\Leftrightarrow x+7+\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)=0\)
=>x+7+1/3-1/10=0
hay x=-217/30
a) Liệt kê
x = {-7;-6;-5;-4;-3;-2;-1;0;1;2;3;4;5;6;7}
Tính tổng là: -7+-6+-5+-4+.....+4+5+6+7
= (-7+7)+(-6+6)+(-5+5)+....+(-1+1)+0
= 0+0+0....+0
= 0
b) Liệt kê
x = {-5;-4;-3;-2;-1;0;1;2;3}
Tính tổng: -5+-4+-3+-2+-2+0+1+2+3
= (-3+3)+(-2+2)+(-1+1)+0+-5+-4
= 0+0+0+0+ -9
= -9
c) Liệt kê:
x = { -19;-18;-17;-16;....;18;19;20}
Tính tổng: -19+-18+-17+-16+....+15+16+17+18+19+20
= (-19+19)+(-18+18)+...+(-1+1)+0+20
= 0 + 0+...+0+20
= 20
*TÌM X:
a) 2x -35 = 15
2x = 15 + 35
2x = 50
x = 50 :2
x = 25
b) 3x + 17 = 2
3x = 17+2
3x = 19
x = 19 : 3
x = 6,33
c) /x-1/ = 0
\(\hept{\begin{cases}x-1=0\\x-1=-0\left(loai\right)\end{cases}}\)
Vậy x-1 = 0
x = 0 +1 = 1
1)Ta có: \(12,\left(1\right)=12+0,\left(1\right)=12+\frac{1}{9}=\frac{109}{9}\);
\(2,3\left(6\right)=2,3+\frac{1}{10}\times0,\left(6\right)=2,3+\frac{1}{10}\times6\times0,\left(1\right)=2,3+\frac{1}{10}\times6\times\frac{1}{9}=\frac{71}{30}\)\(4,\left(21\right)=4+21\times0,\left(01\right)=4+21\times\frac{1}{99}=\frac{139}{33}\)
\(\Rightarrow\)\(\left[\frac{109}{9}-\frac{71}{30}\right]\div\frac{139}{33}=\frac{9647}{4170}\)
2)Ta có: \(0,\left(12\right)=12\times0,\left(01\right)=12\times\frac{1}{99}=\frac{4}{33}\)
\(1,\left(6\right)=1+6\times0,\left(1\right)=1+6\times\frac{1}{9}=\frac{5}{3}\)
\(0,\left(4\right)=4\times0,\left(1\right)=4\times\frac{1}{9}=\frac{4}{9}\)
\(\Rightarrow\frac{4}{33}\div\frac{5}{3}=x\div\frac{4}{9}\Rightarrow x\div\frac{4}{9}=\frac{4}{55}\Rightarrow x=\frac{4}{55}\times\frac{4}{9}\Rightarrow x=\frac{16}{495}\)
\(\dfrac{1}{2}-3x+\left|x-1\right|=0\\ \Rightarrow3x+\left|x-1\right|=\dfrac{1}{2}-0\\ \Rightarrow3x+\left|x-1\right|=\dfrac{1}{2}\\ \Rightarrow\left|x-1\right|=\dfrac{1}{2}-3x\\ \Rightarrow\left[{}\begin{matrix}x-1=\dfrac{1}{2}-3x\\x-1=-\dfrac{1}{2}+3x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x+3x=\dfrac{1}{2}+1\\x-3x=-\dfrac{1}{2}+1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}4x=\dfrac{3}{2}\\2x=\dfrac{1}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{8}\\x=\dfrac{1}{4}\end{matrix}\right.\)
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\(\dfrac{1}{2}\left|2x-1\right|+\left|2x-1\right|=x+1\\ \Rightarrow\left|2x-1\right|\cdot\left(\dfrac{1}{2}+1\right)=x+1\\ \Rightarrow\left|2x-1\right|\cdot\dfrac{3}{2}=x+1\\ \Rightarrow\left|2x-1\right|=x+1:\dfrac{3}{2}\\ \Rightarrow\left|2x-1\right|=x+\dfrac{2}{3}\\ \Rightarrow\left[{}\begin{matrix}2x-1=x+\dfrac{2}{3}\\2x-1=-x-\dfrac{2}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x-x=\dfrac{2}{3}+1\\2x+x=-\dfrac{2}{3}+1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\3x=\dfrac{1}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=\dfrac{1}{9}\end{matrix}\right.\)
kohieeur gì sất
theo mình thì cậu nên hỏi cô giáo là tốt nhất
hoặc ghi bó tay vào đó
cậu ơi đè cậu là :
x + ( x + 1 ) + ( x + 2 ) + .... +(x+ 19 )+ (x+20) +(x+ 21) = 0
hay thế này : x + ( x + 1 ) + ( x + 2 ) + 3+.... + 19 + 20 + 21 = 0 ?
là x + ( x + 1 ) + ( x + 2 ) + ..... + 19 + 20 + 21 = 0
nha bạn