2+4/3x1/6=
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a) \(\dfrac{11}{3}-\dfrac{2}{3}\times\dfrac{5}{6}\)
\(=\dfrac{11}{3}-\dfrac{10}{18}\)
\(=\dfrac{11}{3}-\dfrac{5}{9}\)
\(=\dfrac{33}{9}-\dfrac{5}{9}\)
\(=\dfrac{28}{9}\)
b) \(\dfrac{13}{2}+\dfrac{7}{6}:\dfrac{2}{3}\)
\(=\dfrac{13}{2}+\dfrac{7}{6}\cdot\dfrac{3}{2}\)
\(=\dfrac{13}{2}+\dfrac{7}{4}\)
\(=\dfrac{26}{4}+\dfrac{7}{4}\)
\(=\dfrac{33}{4}\)
c) \(\dfrac{28}{9}-\dfrac{3}{4}+\dfrac{1}{3}\)
\(=\dfrac{28}{9}+\dfrac{3}{9}-\dfrac{3}{4}\)
\(=\dfrac{31}{9}-\dfrac{3}{4}\)
\(=\dfrac{124}{36}-\dfrac{27}{36}\)
\(=\dfrac{97}{36}\)
d) \(\dfrac{5}{8}:\dfrac{2}{3}\times\dfrac{1}{5}\)
\(=\dfrac{5}{8}\times\dfrac{3}{2}\times\dfrac{1}{5}\)
\(=\dfrac{15}{80}\)
\(=\dfrac{3}{16}\)
a) \(\dfrac{11}{3}-\dfrac{2}{3}\times\dfrac{5}{6}=\dfrac{11}{3}-\dfrac{5}{9}=\dfrac{28}{9}\)
b) \(\dfrac{13}{2}+\dfrac{7}{6}:\dfrac{2}{3}=\dfrac{13}{2}+\dfrac{7}{6}\times\dfrac{3}{2}=\dfrac{13}{2}+\dfrac{7}{4}=\dfrac{33}{4}\)
c) \(\dfrac{28}{9}-\dfrac{3}{4}+\dfrac{1}{3}=\dfrac{97}{36}\)
d) \(\dfrac{5}{8}:\dfrac{2}{3}\times\dfrac{1}{5}=\dfrac{5}{8}\times\dfrac{3}{2}\times\dfrac{1}{5}=\dfrac{3}{16}\)
\(x^2-2x-m^2+m-4=0\left(1\right)\)
Để phương trình (1) có 2 nghiệm phân biệt thì:
\(\Delta>0\Rightarrow\left(-2\right)^2-4.\left(-m^2+m-4\right)>0\)
\(\Rightarrow4+4m^2-4m+16>0\)
\(\Leftrightarrow\left(2m-1\right)^2+19>0\) (luôn đúng)
Vậy với \(\forall m\) thì phương trình (1) luôn có 2 nghiệm phân biệt.
Theo định lí Viete cho phương trình (1) ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-2\\x_1x_2=-m^2+m-4\end{matrix}\right.\)
Ta có: \(\left|3x_1\right|-\left|x_2\right|=6\left(2\right)\)
Ta thấy:\(-m^2+m-4=-\left(m^2-m+\dfrac{1}{4}\right)-\dfrac{15}{4}=-\left(m-\dfrac{1}{2}\right)^2-\dfrac{15}{4}\le-\dfrac{15}{4}< 0\)
\(\Rightarrow-m^2+m-4< 0\) hay \(x_1x_2< 0\). Do đó x1, x2 phải trái dấu.
Ta xét 2 trường hợp:
TH1, x1>0 , x2<0. Khi đó:
\(\left(2\right)\Rightarrow3x_1+x_2=6\)
\(\Rightarrow\left(x_1+x_2\right)-6=-2x_1\left(1'\right)\) và \(3\left(x_1+x_2\right)-6=2x_2\left(2'\right)\)
Lấy (1') nhân cho (2') ta được:
\(\left[\left(x_1+x_2\right)-6\right]\left[3\left(x_1+x_2\right)-6\right]=-4x_1x_2\)
\(\Rightarrow\left(-2-6\right)\left[3.\left(-2\right)-6\right]=-4\left(-m^2+m-4\right)\)
\(\Leftrightarrow-m^2+m-4=-24\)
\(\Leftrightarrow m^2-m+4=24\)
\(\Leftrightarrow m^2-m-20=0\)
\(\Leftrightarrow\left[{}\begin{matrix}m=5\\m=-4\end{matrix}\right.\)
TH2: x1<0 ; x2>0. Khi đó:
\(\left(2\right)\Rightarrow3x_1+x_2=-6\)
\(\Rightarrow\left(x_1+x_2\right)+6=-2x_1\left(3'\right)\) và \(3\left(x_1+x_2\right)+6=2x_2\left(4'\right)\)
Lấy (3') nhân cho (4') ta được:
\(\left[\left(x_1+x_2\right)+6\right]\left[3\left(x_1+x_2\right)+6\right]=-4x_1x_2\)
\(\Rightarrow\left(-2+6\right)\left[3.\left(-2\right)+6\right]=-4\left(-m^2+m-4\right)\)
\(\Rightarrow m^2-m+4=0\) (phương trình vô nghiệm)
Thử lại ta có \(\left[{}\begin{matrix}m=5\\m=-4\end{matrix}\right.\)
a) 2/9 : 2/3 x 1/2
= 2/9 x 3/2 x 1/2
= 6/18 x 1/2
= 6/36
b) 2 + 1/4 x 4/3
= 2 + 4/7
= 2/1(7) + 4/7
= 14/7 + 4/7
= 18/7
c) 3 x 1/2 x 1/4
= 3/1 x 1/2 x 1/4
= 3/2 x 1/4
= 3/8
#Hok_tốt
\(\frac{1}{2}\times\frac{1}{3}+\frac{1}{3}\times\frac{1}{4}+...+\frac{1}{8}\times\frac{1}{9}=\frac{1}{2\times3}+\frac{1}{3\times4}+....+\frac{1}{8\times9}\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}=\frac{1}{2}-\frac{1}{9}=\frac{7}{18}\)
2+4/3x1/6
= 2 + 2/9
= 20/9
2 + 4/3 x 1/6
= 2 + 2/9
= 18/9 + 2/9
= 20/9