(-2)7 : (-2)4+ 8
ai giúp mik vs
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\(\dfrac{4}{5}+\dfrac{19}{18}-\dfrac{1}{2}+\dfrac{1}{5}-\dfrac{10}{8}\)
\(=\left(\dfrac{4}{5}+\dfrac{1}{5}\right)-\left(\dfrac{4}{8}+\dfrac{10}{8}\right)+\dfrac{19}{18}\)
\(=\dfrac{5}{5}-\dfrac{14}{8}+\dfrac{19}{18}\)
\(=1-\dfrac{7}{4}+\dfrac{19}{18}\)
\(=-\dfrac{3}{4}+\dfrac{19}{18}=\dfrac{11}{36}\)
\(\dfrac{4}{5}+\dfrac{19}{18}-\dfrac{1}{2}+\dfrac{1}{5}-\dfrac{10}{8}=\dfrac{4}{5}+\dfrac{19}{18}-\dfrac{1}{2}+\dfrac{1}{5}-\dfrac{5}{4}=\left(\dfrac{4}{5}+\dfrac{1}{5}\right)+\left(\dfrac{19}{18}-\dfrac{1}{2}\right)-\dfrac{5}{4}=1+\dfrac{5}{9}-\dfrac{5}{4}=\dfrac{36}{36}+\dfrac{20}{36}-\dfrac{45}{36}=\dfrac{11}{36}\)
\(=6+\dfrac{4}{9}+\dfrac{7}{11}-4-\dfrac{4}{9}+2+\dfrac{4}{11}=4+1=5\)
\(\left(x+7\right)\left(x-4\right)=2\left(x-4\right)\)
\(\Leftrightarrow\left(x+7\right)\left(x-4\right)-2\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+7-2\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x+5=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=4\\x=-5\end{cases}}\)
Vậy : \(x\in\left\{4,-5\right\}\)
\(\left(x+7\right)\left(x-4\right)=2\left(x-4\right)\)
\(\Leftrightarrow x^2-4x+7x-28=2x-8\)
\(\Leftrightarrow x^2+3x-28=2x-8\)
\(\Leftrightarrow x^2+3x-28-2x+8=0\)
\(\Leftrightarrow x^2+x-20=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=-5\end{cases}}}\)
Vậy \(x\in\left\{4;-5\right\}\)
3.42+(57:56)-(2.24)
=3.42+57-6-24+1
=3.42+51-25
=(3.42)+5-32
=48+5-32
=53-32
=21
ĐKXĐ: \(a^3+a^2-3\ge0\) (1)
Đặt \(\sqrt{a^3+a^2-3}=x\ge0\Rightarrow a^3+a^2+4=x^2+7\)
Phương trình trở thành:
\(\sqrt{x^2+7}+x=7\)
\(\Leftrightarrow\sqrt{x^2+7}=7-x\) (\(x\le7\))
\(\Leftrightarrow x^2+7=x^2-14x+49\)
\(\Rightarrow14x=42\)
\(\Rightarrow x=3\)
\(\Rightarrow a^3+a^2-3=9\)
\(\Rightarrow a^3+a^2-12=0\)
\(\Rightarrow\left(a-2\right)\left(a^2+3a+6\right)=0\)
\(\Rightarrow a=2\) (thỏa mãn (1))
\(|\dfrac{4}{3}x-\dfrac{3}{4}|=\left|-\dfrac{1}{3}\right|.\left|x\right|\Leftrightarrow|\dfrac{4}{3}x-\dfrac{3}{4}|=\dfrac{1}{3}.\left|x\right|\left(1\right)\)
Tìm nghiệm \(\dfrac{4}{3}x-\dfrac{3}{4}=0\Leftrightarrow\dfrac{4}{3}x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{3}{4}.\dfrac{3}{4}\Leftrightarrow x=\dfrac{9}{16}\)
\(x=0\)
Lập bảng xét dấu :
\(x\) \(0\) \(\dfrac{9}{16}\)
\(\left|\dfrac{4}{3}x-\dfrac{3}{4}\right|\) \(-\) \(0\) \(-\) \(0\) \(+\)
\(\left|x\right|\) \(-\) \(0\) \(+\) \(0\) \(+\)
TH1 : \(x< 0\)
\(\left(1\right)\Leftrightarrow-\dfrac{4}{3}x+\dfrac{3}{4}=\dfrac{1}{3}.\left(-x\right)\)
\(\Leftrightarrow-\dfrac{4}{3}x+\dfrac{3}{4}=-\dfrac{1}{3}.x\)
\(\Leftrightarrow\dfrac{4}{3}x-\dfrac{1}{3}x=\dfrac{3}{4}\)
\(\Leftrightarrow x=\dfrac{3}{4}\) (loại vì không thỏa \(x< 0\))
TH2 : \(0\le x\le\dfrac{9}{16}\)
\(\left(1\right)\Leftrightarrow-\dfrac{4}{3}x+\dfrac{3}{4}=\dfrac{1}{3}x\)
\(\Leftrightarrow\dfrac{4}{3}x+\dfrac{1}{3}x=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{5}{3}x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{3}{4}.\dfrac{3}{5}\Leftrightarrow x=\dfrac{9}{20}\) (thỏa điều kiện \(0\le x\le\dfrac{9}{16}\))
TH3 : \(x>\dfrac{9}{16}\)
\(\left(1\right)\Leftrightarrow\dfrac{4}{3}x-\dfrac{3}{4}=\dfrac{1}{3}x\)
\(\Leftrightarrow\dfrac{4}{3}x-\dfrac{1}{3}x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{3}{4}\) (thỏa điều kiện \(x>\dfrac{9}{16}\))
Vậy \(x\in\left\{\dfrac{9}{20};\dfrac{3}{4}\right\}\)
= \(\left(7+7^2+7^3\right)+...+\left(7^{58}+7^{59}+7^{60}\right)\)
= \(7\left(1+7+7^2\right)+...+7^{58}\left(1+7+7^2\right)\)
= \(57.7+...+57.7^{58}\) \(⋮57\)
\(=7\left(1+7+7^2\right)+...+7^{58}\left(1+7+7^2\right)\)
\(=57\cdot\left(1+...+7^{58}\right)⋮57\)
\(\frac{4}{2\cdot7}=\frac{4\cdot5}{2\cdot7\cdot5}=\frac{20}{70}\)
\(\frac{11}{7\cdot5}=\frac{11\cdot2}{7\cdot5\cdot2}=\frac{22}{70}\)
\(\frac{9}{2\cdot5}=\frac{9\cdot7}{2\cdot5\cdot7}=\frac{63}{70}\)
=))
(-2)^7 : (-2)^4 + 8
=(-2)^7-4 +8
=(-2)^3 +8
=-8+8
=0
(-2)7 : ( -2)4 + 8
=(-2)3 + 8
= -8 + 8
= 0
chúc bạn hok tốt !