Tìm giá trị của biểu thức: \(B=x+\dfrac{0,2-0,375+\dfrac{5}{11}}{-0,3+\dfrac{9}{16}-\dfrac{15}{22}}vớix=-\dfrac{1}{3}\)
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Q
19 tháng 6 2017
\(B=x+\dfrac{0,2-0,375+\dfrac{5}{11}}{-0,3+\dfrac{9}{16}-\dfrac{15}{22}}\)
\(=x+\dfrac{0,2-0,375+\dfrac{5}{11}}{-0,3+\dfrac{9}{16}-\dfrac{15}{22}}\)
\(=x+\dfrac{-\dfrac{7}{40}+\dfrac{5}{11}}{-\dfrac{369}{880}}\)
\(=x+\dfrac{\dfrac{123}{440}}{\dfrac{369}{880}}\)
\(=x-\dfrac{2}{3}\)
Thay \(x=-\dfrac{1}{3}\) vào biểu thức B.
Ta có: \(-\dfrac{1}{3}-\dfrac{2}{3}=-\dfrac{3}{3}=-1\)
Vậy giá trị biểu thức B tại \(x=-\dfrac{1}{3}\) là -1.
LD
2
OY
21 tháng 7 2021
\(\dfrac{0,2-0,375+\dfrac{5}{11}}{-0,3+\dfrac{9}{16}-\dfrac{15}{22}}=\dfrac{\dfrac{2}{10}-\dfrac{6}{16}+\dfrac{5}{11}}{-\dfrac{3}{10}+\dfrac{9}{16}-\dfrac{15}{22}}=\dfrac{2.\left(\dfrac{1}{10}-\dfrac{3}{16}+\dfrac{5}{22}\right)}{-3.\left(\dfrac{1}{10}-\dfrac{3}{16}-\dfrac{5}{22}\right)}=\dfrac{-2}{3}\)
OY
21 tháng 7 2021
\(\dfrac{0,2-0,375+\dfrac{5}{11}}{-0,3+\dfrac{9}{16}-\dfrac{15}{22}}=\dfrac{\dfrac{2}{10}-\dfrac{6}{16}+\dfrac{5}{11}}{-\dfrac{3}{10}+\dfrac{9}{16}-\dfrac{15}{22}}=\dfrac{2.\left(\dfrac{1}{10}-\dfrac{3}{16}+\dfrac{5}{22}\right)}{-3.\left(\dfrac{1}{10}-\dfrac{3}{16}-\dfrac{5}{22}\right)=\dfrac{-2}{3}}\)
3 tháng 10 2018
\(B=\dfrac{\dfrac{2}{10}-\dfrac{3}{8}+\dfrac{5}{11}}{\dfrac{-3}{10}+\dfrac{9}{16}-\dfrac{15}{22}}\)\(-\dfrac{1}{3}\)
\(B=\dfrac{\dfrac{2}{10}-\dfrac{6}{16}+\dfrac{10}{22}}{\dfrac{-3}{10}+\dfrac{9}{16}-\dfrac{15}{22}}\)\(-\dfrac{1}{3}\)
\(B=\dfrac{2.\left(\dfrac{1}{10}-\dfrac{3}{16}+\dfrac{5}{22}\right)}{-3.\left(\dfrac{1}{10}-\dfrac{3}{16}+\dfrac{5}{22}\right)}\)\(-\dfrac{1}{3}\)
\(B=\dfrac{-2}{3}-\dfrac{1}{3}=-1\)
6)a) \(\left|\dfrac{5}{3}:x\right|=\left|\dfrac{-1}{6}\right|\)
⇒ \(\left|\dfrac{5}{3}:x\right|=\dfrac{1}{6}\)
⇒ \(\dfrac{5}{3}:x=\dfrac{1}{6}\) hoặc \(\dfrac{5}{3}:x=\dfrac{-1}{6}\)
*TH1 : \(\dfrac{5}{3}:x=\dfrac{1}{6}\)
⇒ \(x=\dfrac{5}{3}:\dfrac{1}{6}=10\)
*TH2 : \(\dfrac{5}{3}:x=\dfrac{-1}{6}\)
⇒ \(x=\dfrac{5}{3}:\dfrac{-1}{6}=-10\)
Vậy \(x\) ∈ \(\left\{10;-10\right\}\)
\(b,\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|-\dfrac{3}{4}=\left|\dfrac{-3}{4}\right|\)
⇒ \(\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|-\dfrac{3}{4}=\dfrac{3}{4}\)
⇒\(\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|=\dfrac{3}{4}+\dfrac{3}{4}=\dfrac{3}{2}\)
⇒ \(\dfrac{3}{4}x-\dfrac{3}{4}=\dfrac{3}{2}\) hoặc \(\dfrac{3}{4}x-\dfrac{3}{4}=\dfrac{-3}{2}\)
TH1 : \(\dfrac{3}{4}x-\dfrac{3}{4}=\dfrac{3}{2}\)
⇒ \(\dfrac{3}{4}x=\dfrac{3}{2}+\dfrac{3}{4}=\dfrac{9}{4}\)
⇒\(x=\dfrac{9}{4}:\dfrac{3}{4}=3\)
TH2 : \(\dfrac{3}{4}x-\dfrac{3}{4}=\dfrac{-3}{2}\)
⇒ \(\dfrac{3}{4}x=\dfrac{-3}{2}+\dfrac{3}{4}=\dfrac{-3}{4}\)
⇒ \(x=\dfrac{-3}{4}:\dfrac{3}{4}=-1\)
Vậy \(x\) ∈ \(\left\{3;1\right\}\)
7 tháng 4 2017
a) (1/7.x-2/7).(-1/5.x-2/5)=0
=> 1/7.x-2/7=0hoặc-1/5.x-2/5=0
*1/7.x-2/7=0
1/7.x=0+2/7
1/7.x=2/7
x=2/7:1/7
x=2
b)1/6.x+1/10.x-4/5.x+1=0
(1/6+1/10-4/5).x+1=0
(1/6+1/10-4/5).x=0-1
(1/6+1/10-4/5).x=-1
(-8/15).x=-1
x=-1:(-8/15) =15/8
19 tháng 9 2021
1) Thay x=16 vào biểu thức ta có:
\(A=\dfrac{\sqrt{x}}{\sqrt{x+3}}=\dfrac{\sqrt{16}}{\sqrt{16}+3}=\dfrac{4}{4+3}=\dfrac{4}{7}\)
2) \(A+B=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\\ \Rightarrow A+B=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{2\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(\Rightarrow A+B=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(\Rightarrow A+B=\dfrac{3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(\Rightarrow A+B=\dfrac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(\Rightarrow A+B=\dfrac{3}{\sqrt{x}+3}\)
19 tháng 9 2021
1: Thay x=16 vào A, ta được:
\(A=\dfrac{4}{4+3}=\dfrac{4}{7}\)
MV
7 tháng 1 2018
\(A=\dfrac{0,375-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{-0,625+0,5-\dfrac{5}{11}-\dfrac{5}{12}}=\dfrac{-3\left(-0,125+0,1-\dfrac{1}{11}-\dfrac{1}{12}\right)}{5\left(-0,125+0,1-\dfrac{1}{11}-\dfrac{1}{12}\right)}=\dfrac{-3}{5}\)
CV
7 tháng 1 2018
A=\(\dfrac{0,375-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{-0,625+0,5-\dfrac{5}{11}-\dfrac{5}{12}}\)
A=\(\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{\dfrac{-5}{8}+\dfrac{5}{10}-\dfrac{5}{11}-\dfrac{5}{12}}\)
A=\(\dfrac{3.\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}{-5.\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}\)
A=\(\dfrac{3}{-5}=\dfrac{-3}{5}\)
\(B=x+\dfrac{0,2-0,375+\dfrac{5}{11}}{-0,3+\dfrac{9}{16}-\dfrac{15}{22}}\)
\(=x+\dfrac{\dfrac{1}{5}-\dfrac{3}{8}+\dfrac{5}{11}}{-\left(\dfrac{3}{10}-\dfrac{9}{16}+\dfrac{15}{22}\right)}\)
\(=x+\dfrac{\dfrac{1}{5}-\dfrac{3}{8}+\dfrac{5}{11}}{-\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{3}{8}+\dfrac{5}{11}\right)}\)
\(=x+\dfrac{1}{-\dfrac{3}{2}}\)
\(=x+\dfrac{-2}{3}\)
Với \(x=-\dfrac{1}{3}\), ta được:
\(B=-\dfrac{1}{3}+\dfrac{-2}{3}=-\dfrac{3}{3}=-1\)
Thanks ạ