\(\frac{x-5}{20}=\frac{3}{4}\)
Bài 1:Tìm x
a) b) c) d)
e)
Bài 2:Tính giá trị
A= B= C= D=
E= F=
Bài 3:Tính
A = a - b + c biết B = a - b - c với
Đáp án: thiếu đề
@#@
mời bn xem xét lại đề bài.
~hok tốt~
\(\frac{2}{x-1.x-3}+\frac{5}{x-3.x-8}+\frac{12}{x-8.x-20}-\frac{1}{x-20}=\frac{-3}{4}\)
tìm x
Có lẽ bạn viết đề sai.
Câu hỏi của Vũ Mai Linh - Toán lớp 7 - Học toán với OnlineMath
Bài 1:Tìm x,biết
a) \(x-\frac{20}{11.13}-\frac{20}{13.15}-\frac{20}{15.17}-...-\frac{20}{53.55}=\frac{3}{11}\)
b)\(x+\frac{15}{90.94}+\frac{15}{94.98}+\frac{15}{98.102}+...+\frac{15}{146.150}=\frac{2}{3}\)
c)\(x-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}=\frac{5}{24}\)
d)\(8-\frac{8-\frac{8}{5}+\frac{8}{25}-\frac{8}{125}}{9-\frac{9}{5}+\frac{9}{25}-\frac{9}{125}}:\frac{161616}{151515}=\frac{4+\frac{4}{73}-\frac{4}{115}}{5+\frac{5}{73}-\frac{1}{23}}\)
a, Câu hỏi của Nguyễn Ánh Ngân - Toán lớp 6 - Học toán với OnlineMath
b, Câu hỏi của Vũ Xuân Hiếu - Toán lớp 6 | Học trực tuyến
c)
câu 1 : \(\frac{x+2}{42}+\frac{x+4}{22}=\frac{x+5}{23}+\frac{x+3}{43}\)
câu 2 : \(\frac{x-10}{20}+\frac{x-20}{10}+\frac{x-30}{5}=\frac{x-14}{4}\)
câu 1 : \(\frac{x+2}{42}+\frac{x+4}{22}=\frac{x+5}{23}+\frac{x+3}{43}\)
câu 2 : \(\frac{x-10}{20}+\frac{x-20}{10}+\frac{x-30}{5}=\frac{x-14}{4}\)
tìm x
\(\frac{x}{x+4}\)= \(\frac{5}{6}\)
\(\frac{x-3}{x+5}\)=\(\frac{5}{7}\)
\(\frac{x+4}{20}\)=\(\frac{5}{20}\)
\(\frac{x+4}{20}\)=\(\frac{5}{x+4}\)
giúp mình vs
\(\frac{x}{x+4}=\frac{5}{6}=>6x=5\left(x+4\right)=5x+20\)
\(=>6x-5x=20=>x=20\)
áp dụng \(\frac{a}{b}=\frac{c}{d}< =>a.d=b.c\)
\(\frac{x+4}{20}=\frac{5}{x+4}\)
\(\Rightarrow\left(x+4\right)^2=20.5=100=10^2\)
\(\Rightarrow\orbr{\begin{cases}x+4=10\\x+4=-10\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=6\\x=-14\end{cases}}\)
\(\frac{11}{40}+\frac{4}{5}-\text{|}\frac{3}{4}-1\frac{5}{12}\text{|}=\frac{3}{20}-x\)
Giải phương trình :
1 ) 5( x - 2 ) = 3x + 10
2 ) x2( x - 5 ) - 4x + 20 = 0
3 ) \(\frac{3x+1}{4}+\frac{8x-21}{20}=\frac{3\left(x+2\right)}{5}-2\)
4 ) \(\frac{3}{4x-20}+\frac{7}{6x+30}=\frac{15}{2x^2-50}\)
1) Ta có: \(5\left(x-2\right)=3x+10\)
\(\Leftrightarrow5x-10-3x-10=0\)
\(\Leftrightarrow2x-20=0\)
\(\Leftrightarrow2\left(x-10\right)=0\)
Vì 2>0
nên x-10=0
hay x=10
Vậy: x=10
2) Ta có: \(x^2\left(x-5\right)-4x+20=0\)
\(\Leftrightarrow x^2\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=2\\x=-2\end{matrix}\right.\)
Vậy: x∈{-2;2;5}
3) Ta có: \(\frac{3x+1}{4}+\frac{8x-21}{20}=\frac{3\left(x+2\right)}{5}-2\)
\(\Leftrightarrow\frac{5\left(3x+1\right)}{20}+\frac{8x-21}{20}-\frac{12\left(x+2\right)}{20}+\frac{40}{20}=0\)
\(\Leftrightarrow15x+5+8x-21-12\left(x+2\right)+40=0\)
\(\Leftrightarrow15x+5-8x-21-12x-24+40=0\)
\(\Leftrightarrow-5x=0\)
hay x=0
Vậy: x=0
4) ĐKXĐ: x≠5; x≠-5
Ta có: \(\frac{3}{4x-20}+\frac{7}{6x+30}=\frac{15}{2x^2-50}\)
\(\Leftrightarrow\frac{3}{4\left(x-5\right)}+\frac{7}{6\left(x+5\right)}-\frac{15}{2\left(x-5\right)\left(x+5\right)}=0\)
\(\Leftrightarrow\frac{9\left(x+5\right)}{12\left(x-5\right)\left(x+5\right)}+\frac{14\left(x-5\right)}{12\left(x+5\right)\left(x-5\right)}-\frac{180}{12\left(x-5\right)\left(x+5\right)}=0\)
\(\Leftrightarrow9x+45+14x-70-180=0\)
\(\Leftrightarrow23x-205=0\)
\(\Leftrightarrow23x=205\)
hay \(x=\frac{205}{23}\)(tm)
Vậy: \(x=\frac{205}{23}\)
\(\frac{x-2}{3}=\frac{4}{5}\)
\(\frac{x+3}{-4}=\frac{5}{20}\)
\(\frac{3}{x-1}=\frac{21}{16}\)
a)(X-2)x5=3x4
X-2=12:5
X=2,4-2
X=0,4
Tìm x:
\(\frac{2}{\left(x-1\right).\left(x-3\right)}+\frac{5}{\left(x-3\right).\left(x-8\right)}+\frac{12}{\left(x-8\right).\left(x-20\right)}-\frac{1}{x-20}=\frac{-3}{4}\)\(\frac{-3}{4}\)
chỉ có một số \(\frac{-3}{4}\) thôi nha
ĐKXXD : \(x\ne20;8;3;1\)
\(\frac{2}{\left(x-1\right)\left(x-3\right)}+\frac{5}{\left(x-3\right)\left(x-8\right)}+\frac{12}{\left(x-8\right)\left(x-20\right)}-\frac{1}{x-20}=-\frac{3}{4}\)
\(\Leftrightarrow\frac{\left(x-1\right)-\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}+\frac{\left(x-3\right)-\left(x-8\right)}{\left(x-3\right)\left(x-8\right)}+\frac{\left(x-8\right)-\left(x-20\right)}{\left(x-8\right)\left(x-20\right)}-\frac{1}{x-20}=-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{x-3}-\frac{1}{x-1}+\frac{1}{x-8}-\frac{1}{x-3}+\frac{1}{x-20}-\frac{1}{x-8}+\frac{1}{x-20}=-\frac{3}{4}\)
\(\Leftrightarrow-\frac{1}{x-1}=-\frac{3}{4}\Leftrightarrow x-1=\frac{4}{3}\Rightarrow x=\frac{7}{3}\)