Tìm x , biết:
\(\frac{x-1}{49}+\frac{x-2}{48}+\frac{x-3}{47}+\frac{x-4}{46}+\frac{x-5}{45}\)=5
Mấy ad giải hộ mình nha ,sắp thi rồi:)
Tìm x biết :
\(\frac{x+1}{49}+\frac{x+2}{48}+\frac{x+3}{47}\frac{x+4}{46}+\frac{x+5}{45}=-5\)
\(\frac{x+1}{49}+1+\frac{x+2}{48}+1+\frac{x+3}{47}+1+\frac{x+4}{46}+1+\frac{x+5}{45}+1=0\)
\(\Leftrightarrow\frac{x+50}{49}+\frac{x+50}{48}+...+\frac{x+50}{45}=0\)
\(\Leftrightarrow\left(x+50\right)\left(\frac{1}{49}+\frac{1}{48}+...+\frac{1}{45}\right)=0\)
Vì 1/49+1/48+...+1/45 khác 0
Nên x+50=0
do đó x=-50
Tim x, biet:
\(\frac{x+1}{49}+\frac{x+2}{48}+\frac{x+3}{47}+\frac{x+4}{46}+\frac{x+5}{45}=-5\)
Ta có :
\(\frac{x+1}{49}+\frac{x+2}{48}+\frac{x+3}{47}+\frac{x+4}{46}+\frac{x+5}{45}=-5\)
\(\Leftrightarrow\)\(\left(\frac{x+1}{49}+1\right)+\left(\frac{x+2}{48}+1\right)+\left(\frac{x+3}{47}+1\right)+\left(\frac{x+4}{46}+1\right)+\left(\frac{x+5}{45}+1\right)=-5+5\)
\(\Leftrightarrow\)\(\frac{x+50}{49}+\frac{x+50}{48}+\frac{x+50}{47}+\frac{x+50}{46}+\frac{x+50}{45}=0\)
\(\Leftrightarrow\)\(\left(x+50\right)\left(\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{46}+\frac{1}{45}\right)=0\)
Vì \(\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{46}+\frac{1}{45}\ne0\)
Nên \(x+50=0\)
\(\Rightarrow\)\(x=-50\)
Vậy \(x=-50\)
Chúc bạn học tốt ~
Tìm \(x\):
a)\(x\) x \(3\frac{1}{3}=3\frac{1}{3}:4\frac{1}{4}\)
b)\(5\frac{2}{3}:x=3\frac{2}{3}-2\frac{1}{2}\)
c)\(\frac{x-1}{49}+\frac{x-2}{48}+\frac{x-3}{47}+\frac{x-4}{46}+\frac{x-5}{45}-5=0\)
Tìm X(X\(\in\)Z):
X+1/49 + X+2/48 + X+3/47 + X+4/46 + X+5/45 = -5
Giải hộ mình nha😊
1. Tìm tất cả x thuộc Z thoả mãn \(\frac{2x+1}{x+3}\) < 0.
2. Số hữu tỉ 43/30 có thể viết được dưới dạng \(1+\frac{1}{x+\frac{1}{y+\frac{1}{z}}}\) Tìm x,y,z
3. Tìm x,y, z biết \(\frac{12x-15y}{7}=\frac{20z-12x}{9}=\frac{15y-20z}{11}\) và x+y+z=48
4. Số bộ (x;y;z) thoả mãn \(xy=\frac{2}{5};yz=\frac{3}{7};xz=-\frac{9}{13}\)và x+y+z=48
Nếu được các bạn ghi cách giải giúp mình nha. Mình sắp thi rồi. Thanks nhiều.
Câu 1: x=-2;-1
Câu 2:
Câu 3: x=20
y=16
z=12
Câu 4: 0 bộ
\(\frac{x}{50}+\frac{x-1}{49}+\frac{x-2}{48}+\frac{x-3}{47}+\frac{x-150}{25}=0\)
Tìm x
Các bạn giải cụ thể từng bước dùm mình cho mình hiểu nha. C.ơn các bạn nhiều
\(\frac{x+43}{57}+\frac{x+46}{54}=\frac{x+49}{51}+\frac{x+52}{48}\)
<=>\(\frac{x+43}{57}+1+\frac{x+46}{54}+1=\frac{x+49}{51}+1+\frac{x+52}{48}+1\)
<=>\(\frac{x+100}{57}+\frac{x+100}{54}=\frac{x+100}{51}+\frac{x+100}{48}\)
<=>\(\frac{x+100}{57}+\frac{x+100}{54}-\frac{x+100}{51}-\frac{x+100}{48}=0\)
<=>\(\left(x+100\right)\left(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\right)=0\)
Vì \(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\ne0\)
=>x+100=0
<=>x=-100
k nha bạn
\(\Leftrightarrow\frac{37x+1648}{1026}=\frac{11x+556}{272}\Rightarrow\left(37x+1648\right)272=1026\left(11x+556\right)\)
<=>(37x+1648)272=272(37x+1648)
=>272(37x+1648)=1026(11x+556)
=>10064x+448256=11286x+570456
<=>-1222x=122200
=>x=122200:-1222
=>x=-100 ( dễ hiểu chưa hả )
\(\frac{x+43}{57}+\frac{x+46}{54}=\frac{x+49}{51}+\frac{x+52}{48}\)
\(\Leftrightarrow\)\(\frac{x+43}{57}+1+\frac{x+46}{54}=\frac{x+49}{51}+1+\frac{x+52}{48}+1\)
\(\Leftrightarrow\)\(\frac{x+43}{57}+\frac{57}{57}+\frac{x+46}{54}+\frac{54}{54}=\frac{x+49}{51}+\frac{51}{51}+\frac{x+52}{48}+\frac{48}{48}\)
\(\Leftrightarrow\)\(\frac{x+100}{57}+\frac{x+100}{54}-\frac{x+100}{51}-\frac{x+100}{48}=0\)
\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\right)=0\)
Vì \(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\ne0\)nên \(x+100=0\Leftrightarrow x=-100\)
Vậy \(x=-100\)
tìm x,biết\(\left(\frac{2}{3}x-\frac{1}{2}\right).\frac{3}{4}-\frac{2}{5}x=\frac{17}{4}\)mấy bạn giải ra luôn hộ mình
\(\Leftrightarrow\frac{1}{2}x-\frac{3}{8}-\frac{2}{5}x=\frac{17}{4}\)
\(\Leftrightarrow\frac{1}{2}x-\frac{2}{5}x=\frac{17}{4}+\frac{3}{8}\)(Bạn tự quy đồng chỗ này)
\(\Leftrightarrow\frac{1}{10}x=\frac{37}{8}\)
\(\Leftrightarrow x=\frac{185}{4}\)
a) \(\frac{59-x}{41}+\frac{57-x}{43}=\frac{55-x}{45}+\frac{53+x}{47}+\frac{51-x}{49}=-5\)
b) \(\frac{2-x}{2016}-1=\frac{1-x}{2017}-\frac{x}{2018}\)
Giúp mình với!
a, <=> (59-x/41 + 1) + (57-x/43 + 1) + (55-x/45 + 1) + (53-x/47 + 1) + (51-x/49 + 1) = 0
<=> 100-x/41 + 100-x/43 + 100-x/45 + 100-x/47 + 100-x/49 = 0
<=> (100-x).(1/41+1/43+1/45+1/47+1/49) = 0
<=> 100-x=0 ( vì 1/41+1/43+1/45+1/47+1/49 > 0 )
<=> x=100
Vậy x = 100
b, <=> 2-x/2016 + 1 = (1-x/2017 + 1) + (1 - x/2018)
<=> 2018-x/2016 = 2018-x/2017 + 2018-x/2018
<=> 2018-x/2016 - 2018-x/2017 - 2018-x/2018 = 0
<=> (2018-x).(1/2016-1/2017-1/2018) = 0
<=> 2018-x=0 ( vì 1/2016-1/2017-1/2018 khác 0 )
<=> x=2018
Vậy x=2018
Tk mk nha