\(\frac{69}{91}=\frac{x-5}{133}\)tìm x
Tìm x, biết: \(\frac{x+5}{95}+\frac{x+6}{94}+\frac{x+7}{93}+\frac{x+8}{92}+\frac{x+9}{91}=-5\)
\(\frac{x+5}{95}+\frac{x+6}{94}+\frac{x+7}{93}+\frac{x+8}{92}+\frac{x+9}{91}=-5\)
\(\left(\frac{x+5}{95}+1\right)+\left(\frac{x+6}{94}+1\right)+\left(\frac{x+7}{93}+1\right)+\left(\frac{x+8}{92}+1\right)+\left(\frac{x+9}{91}+1\right)=-5+5=0\)
\(\frac{x+3}{97}+\frac{x+5}{95}+\frac{x+9}{91}=\frac{x+91}{9}+\frac{x+92}{8}+\frac{x+61}{39}\)
Ta có\(\frac{x+3}{97}+\frac{x+5}{95}+\frac{x+9}{91}=\frac{x+91}{9}+\frac{x+92}{8}+\frac{x+61}{39}\)
<=> \(\left(\frac{x+3}{97}+1\right)+\left(\frac{x+5}{95}+1\right)+\left(\frac{x+9}{91}+1\right)=\left(\frac{x+91}{9}+1\right)+\left(\frac{x+92}{8}+1\right)+\left(\frac{x+61}{39}+1\right)\)
<=>\(\frac{x+100}{97}+\frac{x+100}{95}+\frac{x+100}{91}=\frac{x+100}{9}+\frac{x+100}{8}+\frac{x+100}{39}\)
<=>\(\frac{x+100}{97}+\frac{x+100}{95}+\frac{x+100}{91}-\frac{x+100}{9}-\frac{x+100}{8}-\frac{x+100}{39}=0\)
<=> \(\left(x+100\right)\left(\frac{1}{97}+\frac{1}{95}+\frac{1}{91}-\frac{1}{9}-\frac{1}{8}-\frac{1}{39}\right)=0\)
Do \(\frac{1}{97}+\frac{1}{95}+\frac{1}{91}-\frac{1}{9}-\frac{1}{8}-\frac{1}{39}\ne0\)
Nên x+100=0 => x=-100
B=81x[\(\frac{12-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}}{4-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}}:\frac{5+\frac{5}{13}+\frac{5}{169}+\frac{5}{91}}{6+\frac{5}{13}+\frac{5}{169}+\frac{5}{91}}\) ]x\(\frac{158158158}{711711711}\)
Tìm B
\(B=81\cdot\left(\frac{12-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}}{4-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}}:\frac{5+\frac{5}{13}+\frac{5}{169}+\frac{5}{91}}{6+\frac{5}{13}+\frac{5}{169}+\frac{5}{91}}\right)\cdot\frac{158158158}{711711711}\)
\(B=81\cdot\left(\frac{12\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}{4\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}:\frac{5\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}{5\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)+1}\right)\cdot\frac{2}{9}\)
\(B=81\cdot\left(\frac{12}{4}:\frac{6470}{7653}\right)\cdot\frac{2}{9}\)
Xem lại đề bài bẹn owii -.-
BÀI 1:TÌM X
a)\(\frac{X}{108}=\frac{-7}{9}\times\frac{5}{6}\)
b)\(\frac{x+5}{95}+\frac{x+6}{94}+\frac{x+7}{93}+\frac{x+8}{92}+\frac{x+9}{91}=-5\)
a)\(\frac{x}{108}=\frac{-7}{9}.\frac{5}{6}\)
\(\frac{x}{108}=\frac{-35}{54}\)
\(\frac{x}{108}=\frac{-70}{108}\)
\(x=-70\)
b)
Tìm y biết:
65/91=x-5/133
Trả lời: y = ................
\(\frac{65}{91}=x-\frac{5}{133}\)
\(x=\frac{65}{91}+\frac{5}{133}\)
\(x=\frac{100}{133}\)
Bài 1: Tìm x biết \(\frac{5^{x+2}+5^{x+1}+5^x}{31}=\frac{9^{2x}+9^{2x+1}+9^{2\left(x+1\right)}}{91}\)
Bài 18 tìm x
a) x - \(\frac{5}{7}=\frac{1}{9}\)
b) \(\frac{-3}{7}-x=\frac{4}{5}+\frac{-2}{3}\)
c) \(x-\frac{1}{5}=\frac{2}{7}.\frac{-11}{5}\)
d) \(\frac{x}{182}=\frac{-6}{14}.\frac{35}{91}\)
a) \(x-\frac{5}{7}=\frac{1}{9}\Rightarrow x=\frac{1}{9}+\frac{5}{7}\Rightarrow x=\frac{52}{63}\)
b) \(\frac{-3}{7}-x=\frac{4}{5}+\frac{-2}{3}\Rightarrow\frac{-3}{7}-x=\frac{2}{15}\Rightarrow x=\frac{-3}{7}-\frac{2}{15}\Rightarrow x=\frac{-59}{105}\)
c) \(x-\frac{1}{5}=\frac{2}{7}.\frac{-11}{5}\Rightarrow x-\frac{1}{5}=\frac{-22}{35}\Rightarrow x=\frac{-22}{35}+\frac{1}{5}\Rightarrow x=\frac{-3}{7}\)
d) \(\frac{x}{182}=\frac{-6}{14}.\frac{35}{91}\Rightarrow\frac{x}{182}=\frac{-15}{91}\Rightarrow x=\frac{\left(-15\right).182}{91}\Rightarrow x=-30\)
Tìm x, y, z biết :
\(\frac{x}{5}=\frac{y}{4}=\frac{z}{-3}\) và \(x^3-y^3+z^3=69\)
đặt k bn ơi , cần thiết , kết bn vs mk , mk giải rõ cho
a) Tính nhanh:
\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)
b) Tìm x biết:
\(4\times x+69\div x+5\)
a) \(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)
\(=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{13}\right)\)
\(=\frac{1}{2}.\frac{10}{39}\)
\(=\frac{5}{39}\)
a)1/3.5+1/5.7+...+1/11.13
=1/2x(1/3-1/5+1/5-1/7+...+1/11-1/13)
=1/2x(1/3-1/13)
=1/2x10/39
=5/39
quá dễ cái này lớp 4 mik hok rùi thật