1/x(x+1) +1/(x+1)(x+2) +1/(x+2)(x+3) = 3/10. Cho mình câu trả lời dễ hiểu nhé (=.=)
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a,10/3×x-1/2=1/10
10/3×x =1/10+1/2
10/3×x =3/5
=>x =3/5÷10/3
=9/50.
b,1/3×x+1/3×x=1/8
1/3×(x+x) =1/8
x+x = 1/8÷1/3
x+x =3/8
=>x =3/8:2
=3/16.
c,Câu này bạn cần ghi rõ ra mk mới làm được nhé!!!
các bạn ơi nhanh tìm ra đáp án nhé mình đang rất là gấp
a) \(x:\frac{1}{2}+\frac{3}{4}=1\frac{19}{20}\)
\(x:\frac{1}{2}+\frac{3}{4}=\frac{39}{20}\)
\(x:\frac{1}{2}=\frac{39}{20}-\frac{3}{4}\)
\(x:\frac{1}{2}=\frac{39}{20}-\frac{15}{20}\)
\(x:\frac{1}{2}=\frac{24}{20}\)
\(x=\frac{24}{20}.\frac{1}{2}\)
\(x=\frac{3}{5}\)
\(x:\frac{1}{2}+\frac{3}{4}=1\frac{19}{20}\)
\(x:\frac{1}{2}=1\frac{19}{20}-\frac{3}{4}\)
\(x:\frac{1}{2}=\frac{39}{20}-\frac{3}{4}\)
\(x:\frac{1}{2}=\frac{39}{20}-\frac{15}{20}\)
\(x:\frac{1}{2}=\frac{24}{20}\)
\(x=\frac{24}{20}\times\frac{1}{2}\)
\(x=\frac{24}{40}\)
\(x=\frac{3}{5}\)
<=> 4(x^2 + 2x + 1) + 4x^2 - 4x +1 - 8(x^2 - 1) = 11
<=> 4x^2 + 8x + 4 + 4x^2 - 4x +1 - 8x^2 +8 - 11 = 0
<=> 4x + 2 = 0
<=> x = - 1/2
\(\frac{2}{3}\times\frac{1}{5}+\frac{3}{10}\times\frac{2}{3}+\frac{4}{6}\times\frac{1}{2}\)
\(=\frac{2}{3}\times\frac{1}{5}+\frac{3}{10}\times\frac{2}{3}+\frac{2}{3}\times\frac{1}{2}\)
\(=\frac{2}{3}\times\left(\frac{1}{5}+\frac{3}{10}+\frac{1}{2}\right)\)
\(=\frac{2}{3}\times\frac{10}{10}\)
\(=\frac{2}{3}\)
a) \(\frac{2}{5}\)+ X = \(\frac{13}{15}\) - \(\frac{1}{3}\)
\(\frac{2}{5}+X=\frac{8}{15}\)
\(X=\frac{8}{15}-\frac{2}{5}\)
\(X=\frac{2}{15}\)
a) 2/5 + x = 13/15 - 1/3
2/5 + x = 8/15
x = 8/15 - 2/5
x = 2/15
b) x - 2/9 = 2/3 x 2
x - 2/9 = 4/3
x = 4/3 + 2/9
x = 14/9
\(\Leftrightarrow\frac{1}{x^2+5x+6}+\frac{1}{x^2+3x+2}+\frac{1}{x^2+x}=\frac{3}{10}\)
\(\Leftrightarrow\frac{1}{x^2+5x+6}+\frac{1}{x^2+3x+2}+\frac{1}{x^2+x}-\frac{3}{10}=0\)
\(\Leftrightarrow-\frac{3\left(x^2+3x-10\right)}{10x\left(x+3\right)}=0\)
\(\Leftrightarrow3\left(x^2+3x-10\right)=0\)
\(\Leftrightarrow x^2+3x-10=0\)
\(\Leftrightarrow x^2-2x+5x-10=0\)
\(\Leftrightarrow x\left(x-2\right)+5\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow x-2=0\)hoặc\(x+5=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}\)
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}=\frac{3}{10}\)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}=\frac{3}{10}\)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+3}=\frac{3}{10}\Leftrightarrow\frac{\left(x+3\right)-x}{x\left(x+3\right)}=\frac{3}{10}\Leftrightarrow\frac{3}{x\left(x+3\right)}=\frac{3}{10}\)
<=>x(x+3)=10 <=> x2+3x=10 <=> x2+3x-10=0
<=>-(x2-3x+10)=0
<=>x2-3x+10=0
<=>x2-2.x.\(\frac{3}{2}\)+ \(\left(\frac{3}{2}\right)^2+\frac{31}{4}\)=0
<=> \(\left(x-\frac{3}{2}\right)^2+\frac{31}{4}\)=0
Vì \(\left(x-\frac{3}{2}\right)^2\ge0\Rightarrow\left(x-\frac{3}{2}\right)^2+\frac{31}{4}\ge\frac{31}{4}>0\) (với mọi x)
=>PT vô nghiệm