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Ta có :
3x + y - 2x = 14
x : y : z = 3 : 8 : 5 hay \(\frac{x}{3}=\frac{y}{8}=\frac{z}{5}\)
Áp dụng tính chất dãy tỉ số bàng nhau , ta có :
\(\frac{x}{3}=\frac{y}{8}=\frac{z}{5}=\frac{3x}{9}=\frac{2z}{10}=\frac{3x+y-2z}{9+8-10}=\frac{14}{7}=2\)
\(\Rightarrow\hept{\begin{cases}x=2.3=6\\y=2.8=16\\z=2.5=10\end{cases}}\)
TL
Có:
x3=y8=z5=3x9=2z10
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
x3=y8=z5=3x9=2z10=3x+y−2z9+8−10=147=2
x3=2⇒x=6x3=2⇒x=6
y8=2⇒y=16
z5=2⇒z=10
HT
![](https://rs.olm.vn/images/avt/0.png?1311)
a) A = \(9\frac{3}{8}-\left(2\frac{3}{5}+2\frac{3}{8}\right)=9\frac{3}{8}-2\frac{3}{5}-2\frac{3}{8}=\left(9\frac{3}{8}-2\frac{3}{8}\right)-2\frac{3}{5}=7-\frac{13}{5}=\frac{22}{5}\)
b) B = \(\left(15\frac{3}{5}+5\frac{3}{4}\right)-8\frac{3}{5}=15\frac{3}{5}+5\frac{3}{4}-8\frac{3}{5}=\left(15\frac{3}{5}-8\frac{3}{5}\right)+5\frac{3}{4}=7+\frac{23}{4}=\frac{51}{4}\)
c) C = \(17\frac{1}{4}-\left(2\frac{3}{7}+7\frac{1}{4}\right)=17\frac{1}{4}-2\frac{3}{7}-7\frac{1}{4}=\left(17\frac{1}{4}-7\frac{1}{4}\right)-2\frac{3}{7}=10-\frac{17}{7}=\frac{53}{7}\)
d) D = \(\left(11\frac{5}{17}+3\frac{5}{7}\right)-4\frac{5}{17}=11\frac{5}{17}+3\frac{5}{7}-4\frac{5}{17}=\left(11\frac{5}{17}-4\frac{5}{17}\right)+3\frac{5}{7}=7+\frac{26}{7}=\frac{75}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2-4x+5=\left(x-2\right)^2+1\ge0\)
Vậy M(x) không có nghiệm
Vì \(x^2\ge0;4x\ge0\Rightarrow x^2-4x+5\ge0+5>0\)
\(\Rightarrow M\left(x\right)=x^2-4x+5\)không có nghiệm
![](https://rs.olm.vn/images/avt/0.png?1311)
=(11+9+2)+(1/2-3/2+5/2)-(2/3+5/3-7/3)
=22+1,5-0
=23.5
Chúc học tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\frac{1}{2}x+\frac{5}{2}=\frac{7}{2}x-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}x+\frac{5}{2}-\frac{7}{2}x=-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}x-\frac{7}{2}x+\frac{5}{2}=-\frac{3}{4}\)
\(\Leftrightarrow-3x+\frac{5}{2}=-\frac{3}{4}\)
\(\Leftrightarrow-3x=-\frac{13}{4}\)
\(\Leftrightarrow x=-\frac{13}{4}:(-3)=-\frac{13}{4}:\frac{-3}{1}=-\frac{13}{4}\cdot\frac{-1}{3}=\frac{13}{12}\)
\(b,\frac{2}{3}x-\frac{2}{5}=\frac{1}{2}x-\frac{1}{3}\)
\(\Leftrightarrow\frac{2}{3}x-\frac{2}{5}-\frac{1}{2}x=-\frac{1}{3}\)
\(\Leftrightarrow\frac{2}{3}x-\frac{1}{2}x-\frac{2}{5}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{6}x-\frac{2}{5}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{6}x=\frac{1}{15}\)
\(\Leftrightarrow x=\frac{1}{15}:\frac{1}{6}=\frac{1}{15}\cdot6=\frac{6}{15}=\frac{2}{5}\)
\(c,\frac{1}{3}x+\frac{2}{5}(x+1)=0\)
\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)
\(\Leftrightarrow\frac{11}{15}x=-\frac{2}{5}\)
\(\Leftrightarrow x=-\frac{6}{11}\)
d,e,f Tương tự
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(S=\left(5-\frac{2}{3}+\frac{3}{2}\right)-\left(7-\frac{5}{4}-\frac{1}{2}\right)-\left(1-\frac{4}{3}+\frac{2}{5}\right).\)
\(\Rightarrow S=\left(\frac{13}{3}+\frac{3}{2}\right)-\left(\frac{23}{4}-\frac{1}{2}\right)-\left(\frac{-1}{3}+\frac{2}{5}\right)\)
\(\Rightarrow S=\frac{35}{6}-\frac{21}{4}-\frac{1}{15}\)
\(\Rightarrow S=\frac{7}{12}-\frac{1}{15}=\frac{31}{60}\)
Vậy \(S=\frac{31}{60}\)
A = 1 + 52 + 54 + .....+ 5200
25A = 52 + 54 + 56 + .....+ 5200 + 5202
25A - A = 5202 - 1
24A = 5202 - 1
A=1+52+54+...+5200
<=> 25A=52+54+56+........+5202
<=> 25A-A=24A=(5202-1):24