tính gia tri cac bieu thuc sau :
B =(-1/6).(-15/19).(38/45)
D=( 2\(\dfrac{2}{15}\).9/17.3/32): (-3/17)
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a, \(A=\dfrac{1}{3}.\dfrac{-6}{-3}.\dfrac{-9}{10}.\dfrac{-13}{36}\)
\(A=\dfrac{1.\left(-6\right).\left(-9\right).\left(-13\right)}{3.13.10.36}\)
\(A=\dfrac{-1}{10.2}\)
\(A=\dfrac{-1}{20}\)
b, \(B=\dfrac{-1}{3}.\dfrac{-15}{17}.\dfrac{34}{45}\)
\(B=\dfrac{\left(-1\right).\left(-15\right).34}{3.17.45}\)
\(B=\dfrac{2}{3.3}\)
\(B=\dfrac{2}{9}\)
c, \(C=\dfrac{1}{3}.\dfrac{4}{5}+\dfrac{1}{3}.\dfrac{6}{5}+\dfrac{2}{3}\)
\(C=\dfrac{1}{3}.\left(\dfrac{4}{5}+\dfrac{6}{5}\right)+\dfrac{2}{3}\)
\(C=\dfrac{1}{3}.2+\dfrac{2}{3}\)
\(C=\dfrac{2}{3}+\dfrac{2}{3}\)
\(C=\dfrac{4}{3}\)
d, \(D=\dfrac{-5}{6}.\dfrac{4}{19}+\dfrac{-7}{12}.\dfrac{4}{19}-\dfrac{40}{57}\)
\(D=\dfrac{4}{19}.\left(\dfrac{-5}{6}+\dfrac{-7}{12}\right)-\dfrac{40}{57}\)
\(D=\dfrac{4}{19}.\dfrac{-17}{12}-\dfrac{40}{57}\)
\(D=\dfrac{-17}{57}-\dfrac{40}{57}\)
\(D=\dfrac{-57}{57}=-1\)
e, \(E=\dfrac{3}{7}.\dfrac{9}{26}-\dfrac{1}{14}.\dfrac{1}{13}-\dfrac{1}{7}\)
\(E=\dfrac{3}{7}.\dfrac{9}{26}-\left(\dfrac{1}{14}.\dfrac{1}{13}+\dfrac{1}{7}\right)\)
\(E=\dfrac{3}{7}.\dfrac{9}{26}-\left(\dfrac{1}{182}+\dfrac{1}{7}\right)\)
\(E=\dfrac{3}{7}.\dfrac{9}{26}-\dfrac{27}{182}\)
\(E=\dfrac{27}{182}-\dfrac{27}{182}\)
\(E=0\)
a) \(A=3x+15=0\)
\(\Rightarrow3\left(x+5\right)=0\)
\(\Rightarrow x+5=0\)
\(\Rightarrow x=-5\)
b) \(B=2x^2-32=0\)
\(\Rightarrow2\left(x^2-16\right)=0\)
\(\Rightarrow\left(x-4\right)\left(x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-4=0\Rightarrow x=4\\x+4=0\Rightarrow x=-4\end{matrix}\right.\)
b, 53 / 4 . 11/ 16 - 23/4 . 11/16
=11/16(53/4+23/4)
=11/6.19
=209/6
a: \(A=\left|x+1\right|+5\ge5\forall x\)
Dấu '=' xảy ra khi x=-1
b: \(B=\dfrac{x^2+3+12}{x^2+3}=1+\dfrac{12}{x^2+3}\le\dfrac{12}{3}+1=4+1=5\)
Dấu '=' xảy ra khi x=0
a: \(A=\left(2\sqrt{5}-3\sqrt{5}+3\sqrt{5}\right)\cdot\sqrt{5}=2\sqrt{5}\cdot\sqrt{5}=10\)
\(B=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(=\sqrt{x}-1+\sqrt{x}=2\sqrt{x}-1\)
b: A=2B
=>\(10=4\sqrt{x}-2\)
=>\(4\sqrt{x}=12\)
=>x=9(nhận)
a) \(\frac{2}{3}-4.\left(\frac{1}{2}+\frac{3}{4}\right)\)
= \(\frac{2}{3}-4.\frac{5}{4}\)
= \(\frac{2}{3}-5\)
= \(-\frac{13}{3}\).
b) Câu này là 117 hay \(\frac{11}{7}\) vậy bạn?
Chúc bạn học tốt!
\(B=\dfrac{1}{6}\cdot\dfrac{15}{19}\cdot\dfrac{38}{45}=\dfrac{1}{6}\cdot\dfrac{1}{3}\cdot2=\dfrac{1}{9}\)
\(D=\dfrac{32}{15}\cdot\dfrac{9}{17}\cdot\dfrac{3}{32}\cdot\dfrac{-17}{3}=\dfrac{-9}{15}=-\dfrac{3}{5}\)