Tìm x bt:
a)\(\frac{2}{x+7}=\frac{x+7}{32}\)
b)\(\frac{x-2}{x+5}=\frac{x-5}{x+8}\)
GIÚP MK VỚI:3333
Bài 1:Tìm \(x\in Z\)biết
\(A=\frac{3}{x-1}\)\(B=\frac{x-2}{x+3}\)\(C=\frac{2x+1}{x-3}\)
Bài 2:Chứng tỏ rằng:
\(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{63}>2\)\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}< 1\)
Bài 3:Tính hợp lí
\(A=\frac{\frac{3}{5}+\frac{3}{7}-\frac{3}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{11}}\)
\(B=\frac{7}{4}\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)
\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}\)
Ta có : \(\frac{1}{2^2}=\frac{1}{2\cdot2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3^2}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)
...
\(\frac{1}{8^2}=\frac{1}{8\cdot8}< \frac{1}{7\cdot8}\)
Cộng vế theo vế
\(\Rightarrow B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{7\cdot8}\)
\(\Rightarrow B< \frac{1}{1}-\frac{1}{8}=\frac{7}{8}\)
Lại có \(\frac{7}{8}< 1\)
Theo tính chất bắc cầu => \(B< \frac{7}{8}< 1\)
\(\Rightarrow B< 1\left(đpcm\right)\)
Tìm x, biết
a,/\(\frac{13}{11}.\frac{22}{26}-x^2=\frac{7}{16}\)
b,\(^{x^2+\frac{-9}{25}=\frac{2}{5}.\frac{8}{5}}\)
khó quá giúp mk với:))
a/\(\frac{13}{11}.\frac{22}{26}-x^2=\frac{7}{16}\)
\(\Rightarrow1-x^2=\frac{7}{16}\)
\(\Rightarrow x^2=\frac{9}{16}\)
\(\Rightarrow x=\orbr{\begin{cases}\frac{3}{4}\\-\frac{3}{4}\end{cases}}\)
\(a,\frac{13}{11}.\frac{22}{26}-x^2=\frac{7}{16}\)
\(1-x^2=\frac{7}{16}\)
\(x^2=1-\frac{7}{16}=\frac{9}{16}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=-\frac{3}{4}\end{cases}}\)
\(b,x^2+-\frac{9}{25}=\frac{2}{5}.\frac{8}{5}\)
\(x^2+-\frac{9}{25}=\frac{16}{25}\)
\(x^2=\frac{16}{25}--\frac{9}{25}=1\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)
học tốt ~~~
a/ \(\frac{13}{11}.\frac{22}{26}-x^2=\frac{7}{16}\)
\(\Rightarrow\)\(1-x^2=\frac{7}{16}\)
\(\Rightarrow\)\(x^2=1-\frac{7}{16}\)
\(\Rightarrow\)\(x^2=\frac{9}{16}\)
\(\Rightarrow\)\(x^2=\left(\frac{3}{4}\right)^2\)
\(\Rightarrow\)\(x=\frac{3}{4}\)
b, \(x^2+\frac{-9}{25}=\frac{2}{5}.\frac{8}{5}\)
\(\Rightarrow x^2+\frac{-9}{25}=\frac{16}{25}\)
\(\Rightarrow x^2=\frac{16}{25}-\frac{-9}{25}\)
\(\Rightarrow x^2=\frac{25}{25}\)
\(\Rightarrow x^2=\left(\frac{5}{5}\right)^2=\left(\frac{-5}{5}\right)^2\)
\(\Rightarrow x=\frac{5}{5}=\frac{-5}{5}\)
\(\Rightarrow x=1=-1\)
\(\Rightarrow x=\pm1\)
Tìm X ,Y biết ; \(\frac{X+1}{2}=\frac{2.Y-7}{5}=\frac{X-2.Y+8}{X}=\frac{\left(2.Y-7\right)+\left(X-2.Y+8\right)}{5+X}.\) .;;; Các bạn giúp mình với .
Tìm x :
a) \(\frac{x}{5}=\frac{-6}{7}\)
b)\(\frac{x}{2}=\frac{-8}{-x}\)
c) \(x:\frac{1}{x}=\frac{1}{8}:x\)
d) \(\frac{x}{-8}=\frac{x}{-18}\)
e) \(\frac{\left(x-1\right)^2}{3}=\frac{24}{2}\)
giúp mk với nha thank các bn
làm cho 1 cái những cái sau tương tự mà lm nha bạn
\(\frac{x}{5}=-\frac{6}{7}\)
\(=>7x=-6\cdot5\)
\(7x=-30\)
\(x=-\frac{30}{7}\)
\(\frac{x}{2}=-\frac{8}{-x}\)
\(=>\frac{x}{2}=\frac{8}{x}\)
\(=>xx=8\cdot2\)
\(x^2=16\)
\(=>x\in\left\{-4;4\right\}\)
TÍNH
\(\frac{1+0,6-\frac{3}{7}}{\frac{8}{3}+\frac{8}{5}-\frac{8}{7}}-\frac{\frac{1}{3}+0,25-\frac{1}{5}+0,125}{\frac{7}{6}+\frac{7}{8}-0,7+\frac{7}{16}}\)
TÌM X
a) \(\frac{59-x}{41}+\frac{57-x}{43}+\frac{55-x}{45}+\frac{53-x}{47}+\frac{51-x}{49}=-5\)
b) \(\frac{x-5}{1990}+\frac{x-15}{1980}+\frac{x-25}{1970}=\frac{x-1990}{5}+\frac{x-1980}{15}+\frac{x-1570}{25}\)
CÁC BẠN GIÚP MK NHÉ! CẢM ƠN CÁC BẠN NHÌU NHA!
\(\frac{1+0,6-\frac{3}{7}}{\frac{8}{3}+\frac{8}{5}-\frac{8}{7}}=\frac{\frac{3}{3}+\frac{3}{5}-\frac{3}{7}}{\frac{8}{3}+\frac{8}{5}-\frac{8}{7}}=\frac{3.\left(\frac{1}{3}+\frac{1}{5}-\frac{1}{7}\right)}{8.\left(\frac{1}{3}+\frac{1}{5}-\frac{1}{7}\right)}=\frac{3.1}{8.1}=\frac{3}{8}\)
\(\frac{\frac{1}{3}+0,25-\frac{1}{5}+0,125}{\frac{7}{6}+\frac{7}{8}-0,7+\frac{7}{16}}=\frac{\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{8}}{\frac{7}{6}+\frac{7}{8}-\frac{7}{10}+\frac{7}{16}}=\frac{1.\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{8}\right)}{7.\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{8}\right)}=\frac{1.1}{7.1}=\frac{1}{7}\)
=>\(\frac{3}{8}-\frac{1}{7}=\frac{13}{56}\)
Giải các bất phương trình sau:
a) \(\frac{x+2}{98}+\frac{x+3}{97}>\:\frac{x+4}{96}+\frac{x+5}{95}\)
b) \(\frac{x-10}{5}+\frac{x-9}{6}< \frac{x-8}{7}+\frac{x-7}{8}\)
GIÚP MK VỚI, HU HU~~~~~
a) BPT <=> \(\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)>\left(\frac{x+4}{96}+1\right)+\left(\frac{x+5}{95}+1\right)\)
<=> \(\frac{x+100}{98}+\frac{x+100}{97}>\frac{x+100}{96}+\frac{x+100}{95}\)
<=> \(\left(x+100\right)\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)>0\)
Mà \(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}< 0\)
<=> x + 100 < 0
<=> x < -100
b) BPT <=> \(\left(\frac{x-10}{5}-1\right)+\left(\frac{x-9}{6}-1\right)< \left(\frac{x-8}{7}-1\right)+\left(\frac{x-7}{8}-1\right)\)
<=> \(\frac{x-15}{5}+\frac{x-15}{6}< \frac{x-15}{7}+\frac{x-15}{8}\)
<=> \(\left(x-15\right)\left(\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}\right)< 0\)
Mà \(\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}>0\)
<=> x - 15 < 0
<=> x < 15
Các bn giải dùm mk nhé mk rất cần giúp mk với
Bài 2: Tìm X, biết
1) 4+x = \(\frac{x+1}{5}\)
2) \(\frac{7}{X-1}\)= \(\frac{x}{8}\)
a) \(4+x=\frac{x+1}{5}\)
\(5.\left(4+x\right)=x+1\)
\(20+5.x=x+1\)
\(5.x-x=1-20\)
4.x = -19
x = -19/4
2) \(\frac{7}{x-1}=\frac{x}{8}\)
\(x.\left(x-1\right)=7.8\) ( x; x- 1 là 2 số tự nhiên liên tiếp)
=> x = 8
Ai giúp vs !!!
\(a.\frac{3x-7}{5}=\frac{2x-1}{3}\\ b.\frac{4x-7}{12}-x=\frac{3x}{8}\\ c.\frac{x-2009}{1234}+\frac{x-2009}{5678}-\frac{x-2009}{197}=0\\ d.\frac{5x-8}{3}=\frac{1-3x}{2}\\ e.\frac{x-5}{6}-\frac{x-9}{4}=\frac{5x-3}{8}+2\\ f.\frac{x-1}{\frac{2}{5}}-3-\frac{3x-2}{\frac{5}{4}}-2=1\)
\(\frac{3x-7}{5}=\frac{2x-1}{3}\)
\(\Leftrightarrow9x-21=10x-5\)
\(\Leftrightarrow-x=16\Leftrightarrow x=-16\)
\(\frac{4x-7}{12}-x=\frac{3x}{8}\)
\(\Leftrightarrow\frac{4x-7-12x}{12}=\frac{3x}{8}\)
\(\Leftrightarrow\frac{-7-8x}{12}=\frac{3x}{8}\)
\(\Leftrightarrow-56-64x=36x\)
\(\Leftrightarrow-56=100x\Leftrightarrow x=\frac{-14}{25}\)
\(\frac{x-2009}{1234}+\frac{x-2009}{5678}-\frac{x-2009}{197}=0\)
\(\Leftrightarrow\left(x-2019\right)\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)=0\)
Vì \(\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)\ne0\)nên x - 2019 = 0
Vậy x = 2019
\(\frac{5x-8}{3}=\frac{1-3x}{2}\)
\(\Leftrightarrow10x-16=3-9x\)
\(\Leftrightarrow19x=19\Leftrightarrow x=1\)
\(\frac{x-5}{6}-\frac{x-9}{4}=\frac{5x-3}{8}+2\)
\(\Rightarrow\frac{4x-20-6x+54}{24}=\frac{5x-3+16}{8}\)
\(\Rightarrow\frac{-2x+34}{24}=\frac{5x+13}{8}\)
\(\Rightarrow-16x-272=120x+312\)
\(\Leftrightarrow-136x=584\Leftrightarrow x=\frac{-73}{17}\)
ae ơi giúp mk bài mk bài này vs :
tìm x
a) \(\frac{x+3}{x+5}=7\)
b) \(\frac{2x-1}{3x+5}=\frac{-2}{3}\)
c)\(\frac{x+1}{4}=\frac{9}{x+1}\)
d)\(\frac{6x-1}{2x+3}=\frac{3x}{x+2}\)
e) \(\frac{x+1}{x-3}=\frac{x+5}{x-2}\)
f) \(\frac{x+4}{x+5}=\frac{x+6}{x+7}\)
ai nhanh mk tk 10 cái
a)\(\frac{x+3}{x+5}=7\Leftrightarrow x+3=7\left(x+5\right)\)
\(\Leftrightarrow x+3=7x+35\)
\(\Leftrightarrow-6x=32\)
\(\Leftrightarrow x=-\frac{16}{3}\)
b)\(\frac{2x-1}{3x+5}=-\frac{2}{3}\)
\(\Leftrightarrow3\left(2x-1\right)=-2\left(3x+5\right)\)
\(\Leftrightarrow6x-3=-6x-10\)
\(\Leftrightarrow12x=-7\)
\(\Leftrightarrow x=-\frac{7}{12}\)
c)\(\frac{x+1}{4}=\frac{9}{x+1}\Leftrightarrow\left(x+1\right)^2=36\)
\(\Leftrightarrow\left(x+1\right)^2=6^2\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=6\\x+1=-6\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-7\end{cases}}}\)
d)\(\frac{6x-1}{2x+3}=\frac{3x}{x+2}\)
\(\Leftrightarrow\left(6x-1\right)\left(x+2\right)=3x\left(2x+3\right)\)
\(\Leftrightarrow6x^2+12x-x-2=6x^2+9x\)
\(\Leftrightarrow2x=2\Leftrightarrow x=1\)
e)\(\frac{x+1}{x-3}=\frac{x+5}{x-2}\Leftrightarrow\left(x+1\right)\left(x-2\right)=\left(x-3\right)\left(x+5\right)\)
\(\Leftrightarrow x^2-2x+x-2=x^2+5x-3x-15\)
\(\Leftrightarrow-3x=-13\)
\(\Leftrightarrow x=\frac{13}{3}\)
f)\(\frac{x+4}{x+5}=\frac{x+6}{x+7}\Leftrightarrow\left(x+4\right)\left(x+7\right)=\left(x+5\right)\left(x+6\right)\)
\(\Leftrightarrow x^2+11x+28=x^2+11x+30\)
\(\Leftrightarrow0x=2\)
\(\Leftrightarrow x\in\varnothing\)