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Ta có: 9,5. x= 47,4+ 24,8
=> 9,5. x= 72,2
=> x= 72,2: 9,5= 7,6
Vậy x= 7,6
Ta có: x: 8,4+ x: 1,6= 26,3
=> x: \(\frac{42}{5}\)+ x: \(\frac{8}{5}\)= 26,3
=> x. \(\frac{5}{42}\)+ x. \(\frac{5}{8}\)= 26,3
=> x. \(\left(\frac{5}{42}+\frac{5}{8}\right)\)= 26,3
=> x. \(\frac{20+21}{168}\)= 26,3
=> x. \(\frac{41}{168}\)= \(\frac{263}{10}\)
=> x= \(\frac{263}{10}:\frac{41}{168}\)
=> x= \(\frac{263}{10}.\frac{168}{41}\)
=> x\(\approx107,8\)
Vậy x\(\approx107,8\)
Dấu \(\approx\)nghĩa là xấp xỉ nha! Đây là toán lớp 5 nhưng mik làm có hơi xen vào toán lớp 6; 7 một tí. Bạn đọc cho hiểu nha! Có gì ko hiểu thì cứ hỏi mik mik sẽ giải đáp cho!!!!!!!!!!!!!!!!!!!!!!!!!!! Mik mong là bạn sẽ k cho mik
#)Giải :
A, \(\frac{254x399-145}{254+399x253}\)
\(=\frac{253x399+399-145}{254+399x253}\)
\(=\frac{253x399+254}{254+399x253}\)
\(=1\)
B, \(\frac{5932+6001x5931}{5931x6001-69}\)
\(=\frac{5932+6001x5931}{\left(5931+1\right)x6001-69}\)
\(=\frac{5932+6001x5931}{5931x6001+6001-69}\)
\(=\frac{5932+6001x5932}{5932x6001+5932}\)
\(=1\)
#~Will~be~Pens~#
\(\frac{254x399-145}{254+399x253}=\frac{\left(253+1\right)x399-145}{254+399x253}=\frac{253x399+1x399-145}{254+399x253}=\frac{253x399+254}{254+399x253}\)
\(=1\)
a,
\(\frac{326}{325}-\frac{1}{325}=1\) \(\frac{325}{324}-\frac{1}{324}=1\)
Vì \(\frac{1}{325}< \frac{1}{324}\)nên \(\frac{326}{325}< \frac{325}{324}\)
b,
( 11 x 9 - 900 x 0,1 - 8 ) x ( 56,7 x 0,5 + 56,7 x 9,5 )
= ( 99 - 90 - 8 ) x [ 56,7 x ( 0,5 + 9,5 ) ]
= ( 9 - 8 ) x ( 56,7 x 10 )
= 1 x 567 = 567
Ta co : \(A=\frac{399.45+11.5.399}{1995.1996-1991.1995}\)
\(A=\frac{399.45+55.399}{1995.\left(1996-1991\right)}\)
\(A=\frac{399.\left(45+55\right)}{1995.5}\)
\(A=\frac{399.100}{1995.5}\)
\(A=4\)
\(\left(1+x\right)+\left(2+x\right)+\left(3+x\right)+\)\(\left(4+x\right)+\left(5+x\right)=10\times5\)
\(\left(1+2+3+4+5\right)+\left(x+x+x+x+x\right)=50\)
\(15+5x=50\)
\(5x=35\)
\(x=7\)
Vậy \(x=7\)
\(\left(1+x\right)+\left(2+x\right)+\left(3+x\right)+\left(4+x\right)+\left(5+x\right)=10\times5\)
\(\Rightarrow1+x+2+x+3+x+4+x+5+x=50\)
\(\Rightarrow\left(1+2+3+4+5\right)+\left(x+x+x+x+x\right)=50\)
\(\Rightarrow15+5x=50\)
\(\Rightarrow5x=50-15\)
\(\Rightarrow5x=35\)
\(\Rightarrow x=35:5\)
\(\Rightarrow x=7\).
\(x(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{132}=\) \(5\frac{1}{2}\)
\(x\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{11\cdot12}\right)=\frac{11}{2}\)
\(x\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{11}-\frac{1}{12}\right)=\frac{11}{2}\)
\(x\left(1-\frac{1}{12}\right)=\frac{11}{2}\)
\(x\cdot\frac{11}{12}=\frac{11}{2}\)
\(x=\frac{11}{2}:\frac{11}{12}\)
\(x=6\)
Vậy x = 6
\(\frac{x}{2}+\frac{x}{6}+\frac{x}{12}+\frac{x}{20}+...+\frac{x}{132}=5\frac{1}{2}\)
\(\Rightarrow x\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{132}\right)=\frac{11}{2}\)
\(\Rightarrow x\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{11.12}\right)=\frac{11}{2}\)
\(\Rightarrow x\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}=\frac{11}{2}\right)\)
\(\Rightarrow x\left(1-\frac{1}{12}\right)=\frac{11}{2}\)
\(\Rightarrow x.\frac{11}{12}=\frac{11}{12}\)
\(\Rightarrow x=\frac{11}{12}:\frac{11}{12}=1\)
Vậy x = 1
9,5 x X = 399
X = 399 : 9,5
X = 42