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\(A=-1,6:\left(1+\frac{2}{3}\right)\)
\(A=-\frac{16}{10}:\frac{5}{3}\)
\(A=-\frac{16.3}{10.5}=-\frac{48}{50}=-\frac{24}{25}\)
\(B=1,4\times\frac{15}{49}-\left(\frac{4}{5}+\frac{2}{3}\right):2\frac{1}{5}\)
\(B=\frac{14}{10}\times\frac{15}{49}-\left(\frac{4}{5}+\frac{2}{3}\right):\frac{11}{5}\)
\(B=\frac{2.7.3.5}{2.5.7.7}-\left(\frac{12+10}{15}\right):\frac{11}{5}\)
\(B=\frac{3}{7}-\frac{22}{15}:\frac{11}{5}\)
\(B=\frac{3}{7}-\frac{22}{15}\times\frac{5}{11}=\frac{3}{7}-\frac{2.11.5}{3.5.11}\)
\(B=\frac{3}{7}-\frac{2}{3}=\frac{9-14}{21}=-\frac{5}{21}\)
Ủng hộ mk nha !!! ^_^
Bài 1:
a: \(A=\dfrac{\left(85+\dfrac{7}{30}-83-\dfrac{5}{18}\right):\dfrac{8}{3}}{\dfrac{1}{25}}\)
\(=\left(2+\dfrac{7}{30}-\dfrac{5}{18}\right)\cdot\dfrac{3}{8}\cdot25\)
\(=\dfrac{180+21-25}{90}\cdot\dfrac{75}{8}\)
\(=\dfrac{176}{90}\cdot\dfrac{75}{8}=\dfrac{55}{3}\)
=>12,5% của A là 55/8x1/8=55/64
b: \(B=\dfrac{\left(6+\dfrac{3}{5}-3-\dfrac{3}{14}\right)\cdot\dfrac{36}{5}}{19.75:2.5}\)
\(=\dfrac{\left(3+\dfrac{27}{70}\right)\cdot\dfrac{36}{5}}{\dfrac{79}{10}}\)
\(=\dfrac{\dfrac{210+27}{70}\cdot\dfrac{36}{5}}{\dfrac{79}{10}}\)
\(=\dfrac{4266}{175}\cdot\dfrac{10}{79}=\dfrac{108}{35}\)
=>5% là 108/35x1/20=27/175
1 3/7-4/5=10/7-4/5=50/35-28/35=22/35
6/13x-3/10+2/5x4/13
=-9/65+8/65
=-1/65
câu đầu mik tính ra số to mà cx ko chắc là đúng nên mik ko viết
*chúc bn học tốt đạt nhiều điểm cao*
bn đổi ngược hai vế cho nhau là ra 1 bài toán bình thường thôi
Bài nỳ tuy rất dài nhưng cũng dễ
Chí cần cậu cuyển VT sang VP rồi tìm x bình thường
Chúc cậu học tốt
a.15-(5-2x)=-4
\(\Leftrightarrow\)15-5+2x=-4
\(\Leftrightarrow\)2x=-4-15+5
\(\Leftrightarrow\)2x=-14
\(\Leftrightarrow\)x=-7
b.TH1:x-3\(\ge\)0 \(\Rightarrow\)x\(\ge\)3
Ta có \(|\)x-3\(|\)=x-3
PT trên\(\Leftrightarrow\)x-3+1=4
\(\Leftrightarrow\)x=4+3-1
\(\Leftrightarrow\)x=6(nhận)
TH2:x-3<0\(\Leftrightarrow\)x<3
Ta có:\(|\)x-3\(|\)=-x+3
PT trên\(\Leftrightarrow\)-x+3+1=4
-x=4-3-1
x=0(nhận)
Vậy S={0;6}
ta có : \(\dfrac{3}{2}\)A= \(\dfrac{3}{4}+\)\(\left(\dfrac{3}{2}\right)^2+\left(\dfrac{3}{2}\right)^3+\)\(...+\left(\dfrac{3}{2}\right)^{2013}\) (1)
A= \(\dfrac{1}{2}+\dfrac{3}{2}\)\(+\left(\dfrac{3}{2}\right)^2+...+\)\(\left(\dfrac{3}{2}\right)^{2012}\) (2)
Lấy (1) trừ đi (2) vế theo vế:
\(\dfrac{3}{2}A-A=\dfrac{3}{4}-\dfrac{1}{2}-\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^{2013}\)
\(\dfrac{1}{2}A=\left(\dfrac{3}{2}\right)^{2013}-\dfrac{5}{4}\Rightarrow A=\dfrac{3^{2013}}{2^{2012}}-\dfrac{5}{2}\)
ta có : \(B=\left(\dfrac{3}{2}\right)^{2013}:2=\dfrac{3^{2013}}{2^{2013}}.\dfrac{1}{2}=\dfrac{3^{2013}}{2^{2014}}\)
Vậy \(A-B=\dfrac{3^{2013}}{2^{2014}}-\left(\dfrac{3^{2013}}{2^{2012}}-\dfrac{5}{2}\right)\)
\(A=\frac{1}{2}+\frac{3}{2}+\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^3+...+\left(\frac{3}{2}\right)^{2012}\)(1)
\(\frac{3}{2}A=\frac{3}{4}+\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^3+\left(\frac{3}{2}\right)^4+...+\left(\frac{3}{2}\right)^{2013}\)(2)
Lấy (2) trừ (1) ta được:
\(\frac{1}{2}A=\frac{3}{4}+\left(\frac{3}{2}\right)^{2013}-\frac{1}{2}-\frac{3}{2}=\left(\frac{3}{2}\right)^{2013}-\frac{5}{4}\)
\(A=\frac{\left(\frac{3}{2}\right)^{2013}-\frac{5}{4}}{\frac{1}{2}}=\left(\frac{3}{2}\right)^{2013}.2-\frac{5}{4}.2=\left(\frac{3}{2}\right)^{2013}.2-\frac{5}{2}\)
\(\Rightarrow B-A=\left(\frac{3}{2}\right)^{2013}\cdot\frac{1}{2}-\left(\frac{3}{2}\right)^{2013}.2+\frac{5}{2}=-\left(\frac{3}{2}\right)^{2014}+\frac{5}{2}\)
\(6\div2\left(1+2\right)\)
\(=6\div2.3\)
\(=3.3\)
\(=9\)
\(6\div2\left(1+2\right)=6\div2.3\\ =6\div6\\ =1 \)
\(6\div2\times\left(1+2\right)=3\times\left(1+2\right)\\ =3\times3\\ =9\)