Tính gọn biểu thức:
1/ -5.7^5+7^4:7^6.10-2.7^5
2/10^3+5.10^2+5^3:-13
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NM
0
30 tháng 6 2018
a) Với \(2x+1\ge0\Rightarrow x\ge\frac{-1}{2}\) thì \(2x+1-2=1\Rightarrow2x+1=3\Rightarrow2x=2\Rightarrow x=1\) (nhận)
Với 2x + 1 <0 => x < -1/2 thì \(-\left(2x+1\right)-2=1\Rightarrow-2x-1-2=1\Rightarrow-2x=4\Rightarrow x=-2\) (nhận)
Vậy x = {-2;1}
b) Với \(3x-1\ge0\Rightarrow x\ge\frac{1}{3}\) thì \(3x-1=1-3x\Rightarrow6x=2\Rightarrow x=\frac{1}{3}\) (nhận)
Với 3x - 1 < 0 => x < 1/3 thì \(-\left(3x-1\right)=1-3x\Rightarrow-3x+1=1-3x\Rightarrow0x=0\) (nhận)
Vậy x = 1/3 hoặc x vô số nghiệm với x < 1/3
30 tháng 6 2018
a) | 2x + 1| - 2 = 1
| 2x + 1| = 3
TH1: 2x + 1 = 3
2x = 2
x = 1
TH2: 2x + 1 = -3
2x = -4
x = -2
KL: x = 1 hoặc x = - 2
SM
1
7 tháng 8 2019
b) \(\frac{-5\cdot7^5+7^4}{7^6\cdot10-2\cdot7^5}\)
\(=\frac{-35\cdot7^4+7^4}{7^5\cdot70-2\cdot7^5}\)
\(=\frac{7^4\left(-35+1\right)}{7^5\left(70-2\right)}\)
\(=\frac{7^4\cdot\left(-34\right)}{7^5\cdot68}\)
\(=\frac{-1}{14}\)
Chắc sai =))
17 tháng 2 2020
\(B=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+\frac{1}{18\cdot19\cdot20}\)
\(B=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+\frac{2}{18\cdot19\cdot20}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{18\cdot19}-\frac{1}{19\cdot20}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{19\cdot20}\right)\)
\(B=\frac{1}{2}\cdot\frac{189}{380}=\frac{189}{760}\)
\(C=\frac{52}{1\cdot6}+\frac{52}{6\cdot11}+\frac{52}{11\cdot16}+...+\frac{52}{31\cdot36}\)
\(C=\frac{52}{5}\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\frac{5}{11\cdot16}+...+\frac{6}{31\cdot36}\right)\)
\(C=\frac{52}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{31}-\frac{1}{36}\right)\)
\(C=\frac{52}{5}\cdot\left(1-\frac{1}{36}\right)\)
\(C=\frac{91}{9}\)
24 tháng 6 2021
a)\(A=\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)
\(=\sqrt[3]{1+3\sqrt{2}+3\sqrt{2^2}+2\sqrt{2}}-\sqrt[3]{2\sqrt{2}-3\sqrt{2^2}+3\sqrt{2}-1}\)
\(=\sqrt[3]{\left(1+\sqrt{2}\right)^3}-\sqrt[.3]{\left(\sqrt{2}-1\right)^3}\)
\(=1+\sqrt{2}-\left(\sqrt{2}-1\right)=2\)
b)\(B=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)
\(\Leftrightarrow B^3=5+2\sqrt{13}+3\sqrt[3]{\left(5+2\sqrt{13}\right)\left(5-2\sqrt{13}\right)}\left(\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5+2\sqrt{13}}\right)+5-2\sqrt{13}\)
\(\Leftrightarrow B^3=10+3.\sqrt[3]{-27}.B\)
\(\Leftrightarrow B^3+9B-10=0\)
\(\Leftrightarrow\left(B-1\right)\left(B^2+B+10\right)=0\)
\(\Leftrightarrow B=1\) (vì \(B^2+B+10>0\))
c)\(C=\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}\)
\(\Leftrightarrow2C=\sqrt[3]{8\sqrt{5}+16}-\sqrt[3]{8\sqrt{5}-16}=\sqrt[3]{1+3\sqrt{5}+3\sqrt{5^2}+5\sqrt{5}}-\sqrt[3]{5\sqrt{5}-3\sqrt{5^2}+3\sqrt{5}-1}\)
\(=\sqrt[3]{\left(1+\sqrt{5}\right)^3}-\sqrt[3]{\left(\sqrt{5}-1\right)^3}\)
\(=1+\sqrt{5}-\left(\sqrt{5}-1\right)\)
\(\Rightarrow C=1\)
d) \(D=\dfrac{10}{\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}}\left(\dfrac{1+\sqrt{2}}{\sqrt{4-2\sqrt{3}}}:\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\right)\)
\(=\dfrac{10\left(\sqrt[3]{3}+\sqrt[3]{2}\right)}{\left(\sqrt[3]{3}+\sqrt[3]{2}\right)\left(\sqrt[3]{9^2}-\sqrt[3]{6}+\sqrt[3]{2^2}\right)}\left(\dfrac{1+\sqrt{2}}{\sqrt{\left(1-\sqrt{3}\right)^2}}.\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\right)\)
\(=\dfrac{10\left(\sqrt[3]{3}+\sqrt[3]{2}\right)}{5}.\dfrac{1+\sqrt{2}}{\left|1-\sqrt{3}\right|}.\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\)
\(=2\left(\sqrt[3]{3}+\sqrt[3]{2}\right).\dfrac{\left(1+\sqrt{2}\right)\left(\sqrt{2}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(=2\left(\sqrt[3]{3}+\sqrt[3]{2}\right).\dfrac{\left(\sqrt{2}\right)^2-1}{\left(\sqrt{3}\right)^2-1}\)
\(=\sqrt[3]{3}+\sqrt[3]{2}\)
Vậy...
15 tháng 4 2022
a: \(=\dfrac{5}{9}\cdot\dfrac{7}{4}+\dfrac{13}{9}\cdot\dfrac{7}{4}=\dfrac{7}{4}\cdot2=\dfrac{7}{2}\)
b: \(=\dfrac{14}{13}\cdot2+\dfrac{3}{13}\cdot2-2\cdot\dfrac{7}{13}=\dfrac{28}{13}+\dfrac{6}{13}-\dfrac{14}{13}=\dfrac{20}{13}\)