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Bài 1:
a) \(\left(\frac{1}{2}\right)^2\) và \(\left(\frac{1}{2}\right)^5\)
Ta có: \(\left(\frac{1}{2}\right)^2=\frac{1}{4}.\)
\(\left(\frac{1}{2}\right)^5=\frac{1}{32}.\)
Vì \(\frac{1}{4}< \frac{1}{32}.\)
=> \(\left(\frac{1}{2}\right)^2< \left(\frac{1}{2}\right)^5.\)
b) \(\left(2,4\right)^3\) và \(\left(2,4\right)^2\)
Ta có: \(\left(2,4\right)^3=13,824.\)
\(\left(2,4\right)^2=5,76.\)
Vì \(13,284>5,76.\)
=> \(\left(2,4\right)^3>\left(2,4\right)^2.\)
c) \(\left(-1\frac{1}{2}\right)^2\) và \(\left(-1\frac{1}{2}\right)^3\)
Ta có: \(\left(-1\frac{1}{2}\right)^2=\left(-\frac{3}{2}\right)^2=\frac{9}{4}.\)
\(\left(-1\frac{1}{2}\right)^3=\left(-\frac{3}{2}\right)^3=-\frac{27}{8}.\)
Vì số dương luôn lớn hơn số âm nên \(\frac{9}{4}>-\frac{27}{8}.\)
=> \(\left(-1\frac{1}{2}\right)^2>\left(-1\frac{1}{2}\right)^3.\)
Chúc bạn học tốt!
Bài 1:
\(a,A=\frac{-25}{28}.0,21=\frac{-25}{28}.\frac{21}{100}=\frac{-25.21}{28.100}=\frac{-1.25.3.7}{4.7.25.4}=\frac{-1.3}{4.4}=\frac{-3}{16}\)
\(b,B=\left(\frac{13}{24}-\frac{29}{30}\right):\left(-10,2\right)=\left(\frac{65}{120}-\frac{116}{120}\right):\frac{-51}{5}=\frac{-51}{120}.\frac{5}{-51}=\frac{-51.5}{120.\left(-51\right)}=\frac{-51.5}{5.24.\left(-51\right)}=\frac{1}{24}\)
Bài 1 :
a) \(-3+\left(-4\right)-\left(-3\right)+\left(2+7-10\right)=-3-4+3+2+7-10=-5\)
b) \(3-\left(-3+2-7\right)+\left(-4\right)=3+3-2+7-4=7\)
c) \(7+\left(-2-3+7\right)-\left(-2\right)=7-2-3+7+2=17\)
d) \(-\left(-3\right)-\left(-2+3-8\right)+\left(-6\right)=3+2-3+8-6=4\)
Bài 2 :
a) \(x^2-2x-\left(3x-2x\right)=x^2-2x-3x+2x=x^2-3x\)
b) \(-\left(x^2+3x^2\right)-\left(-5x^2+3x\right)=-x^2-3x^2+5x^2-3x=x^2-3x\)
c) \(\left(x-y\right)-\left(x+3y+1\right)=x-y-x-3y-1=-4y-1\)
Bài 1:
a, -3+ (-4) - (-3) + (2 + 7 - 10)
= -3 - 4 + 3 + 2 + 7 - 10
= 5 - 10
= -5.
b, 3 - (-3 + 2 - 7) + (-4)
= 3 + 3 - 2 + 7 - 4
= 11 - 4
= 7
c, 7 + (-2 - 3 + 7) - (-2)
= 7 - 2 - 3 + 7 + 2
= 9 + 2
= 11.
d, - (-3) - (-2 + 3 - 8) + (-6)
= 3 + 2 - 3 + 8 - 6
= 10 - 6
= 4.
Mình chỉ làm bài 1 thôi nhé.
Chúc bạn học tốt!
Bài 1:
\(A=\left(x^3.x^3.x^2\right).\left(y.y^4\right).\left(\frac{2}{5}.\frac{-5}{4}\right)\)
\(A=x^8.y^5.\left(-\frac{1}{2}\right)\)
\(B=\left(x^5.x.x^2\right).\left(y^4.y^2.y\right).\left(\frac{-3}{4}.\frac{-8}{9}\right)\)
\(B=x^8.y^7.\frac{2}{3}\)
Bài 2:
\(A=\left(15.x^2.y^3-12.x^2.y^3\right)+\left(11x^3.y^2-8.x^3.y^2\right)+\left(7x^2-12x^2\right)\)
\(A=3.x^2.y^3+2.x^3.y^2-5x^2\)
B tương tự nhé, đáp án là (theo mình)
\(B=\frac{5}{2}.x^5.y+\frac{7}{3}.x.y^4-\frac{1}{4}.x^2.y^3\)
\(B=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{98}+\left(\dfrac{1}{2}\right)^{99}\)
\(\Rightarrow2B=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{97}+\left(\dfrac{1}{2}\right)^{98}\)
\(\Rightarrow2B-B=1-\left(\dfrac{1}{2}\right)^{99}\)
\(B=1-\left(\dfrac{1}{2}\right)^{99}\)
\(2,\)
\(a,\dfrac{45^{10}.2^{10}}{75^{15}}\)
\(=\dfrac{5^{10}.9^{10}.2^{10}}{25^{15}.3^{15}}\)
\(=\dfrac{5^{10}.3^{20}.2^{10}}{5^{30}.3^{15}}\)
\(=\dfrac{5^{10}.3^{15}.\left(3^5.2^{10}\right)}{5^{10}.3^{15}.\left(5^{20}\right)}\)
\(=\dfrac{3^5.2^{10}}{5^{20}}\)
\(b,\dfrac{2^{15}.9^4}{6^3.8^3}\)
\(=\dfrac{2^{15}.3^8}{2^3.3^3.2^9}=\dfrac{2^{15}.3^8}{2^{12}.3^3}=2^3.3^5\)
\(c,\dfrac{8^{10}+4^{10}}{8^4+4^{11}}=\dfrac{4^{10}.2^{10}+4^{10}}{4^4.2^4+4^4.4^7}=\dfrac{4^4.\left(4^6.2^{10}+4^6\right)}{4^4.\left(2^4+4^7\right)}\)
\(=\dfrac{4^{11}+4^6}{4^8.4^7}=\dfrac{4^6.\left(4^5+1\right)}{4^6.\left(4^2-4\right)}=\dfrac{1024+1}{16-4}=\dfrac{1025}{12}\)
\(d,\dfrac{81^{11}.3^{17}}{27^{10}.9^{15}}=\dfrac{3^{44}.3^{17}}{3^{30}.3^{30}}=\dfrac{3^{61}}{3^{60}}=3\)
\(3,\)
\(a,\left(2x+4\right)^2=\dfrac{1}{4}\)
\(\left(2x+4\right)^2=\left(\dfrac{1}{2}\right)^2=\left(\dfrac{-1}{2}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}2x+4=\dfrac{1}{2}\\2x+4=\dfrac{-1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{1}{2}-4=\dfrac{-7}{2}\\2x=\dfrac{-1}{2}-4=\dfrac{-9}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-7}{4}\\x=\dfrac{-9}{4}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{-7}{4};\dfrac{-9}{4}\right\}\)
\(b,\left(2x-3\right)^2=36\)
\(\left(2x-3\right)^2=6^2=\left(-6\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=6+3=9\\2x=-6+3=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=\dfrac{-3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{9}{2};\dfrac{-3}{2}\right\}\)
\(c,5^{x+2}=628\)
\(5^{x+2}=5^4\)
\(\Rightarrow x+2=4\)
\(\Rightarrow x=4-2=2\)
Vậy \(x=2\)
\(d,\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)
\(\Rightarrow\left(x-1\right)^{x+4}-\left(x-1\right)^{x+2}=0\)
\(\Rightarrow\left(x-1\right)^{x+2}.\left[\left(x-1\right)^2-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^{x+2}=0\\\left(x-1\right)^2-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x-1=1\\x-1=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)
Vậy \(x\in\left\{0;1;2\right\}\)
Bài 1:
B= \(\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{99}\)
2B= \(2.[\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{99}]\)
2B= \(1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{98}\)
⇒2B-B= \(1-\left(\dfrac{1}{2}\right)^{99}\)
B= 1
Vậy B=1
Bài 2:
a, \(\dfrac{45^{10}.2^{10}}{75^{15}}\)= \(\dfrac{\left(3^2.5\right)^{10}.2^{10}}{\left(3.5^2\right)^{15}}=\dfrac{3^{20}.5^{10}.2^{10}}{3^{15}.5^{30}}=\dfrac{3^5.2^{10}}{5^{20}}\)
b, \(\dfrac{2^{15}.9^4}{6^3.8^3}=\dfrac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^3.\left(2^3\right)^3}=\dfrac{2^{15}.3^8}{2^3.3^3.2^9}=\dfrac{2^{15}.3^8}{2^{12}.3^3}=2^3.3^5\)
c,\(\dfrac{8^{10}+4^{10}}{8^4+4^{11}}=\dfrac{\left(2.4\right)^{10}+4^{10}}{\left(2.4\right)^4+4^{11}}=\dfrac{2^{10}.4^{10}+4^{10}}{2^4.4^4+4^{11}}=\dfrac{4^{10}.\left(2^{10}+1\right)}{4^6+4^6.4^5}=\dfrac{4^{10}.\left(2^{10}+1\right)}{4^6.\left(4^5+1\right)}=\dfrac{4^{10}.\left(2^{10}+1\right)}{4^6.\left(2^{10}+1\right)}=4^4=256\)
d, \(\dfrac{81^{11}.3^{17}}{27^{10}.9^{15}}=\dfrac{\left(3^4\right)^{11}.3^{17}}{\left(3^3\right)^{10}.\left(3^2\right)^{15}}=\dfrac{3^{44}.3^{17}}{3^{30}.3^{30}}=\dfrac{3^{61}}{3^{60}}=3\)
Bài 3:
a, \(\left(2x+4\right)^2=\dfrac{1}{4}\)
\(\left(2x+4\right)^2=\left(\dfrac{1}{2}\right)^2\)
\(2x+4=\dfrac{1}{2}\)
\(2x=\dfrac{1}{2}-4\)
\(2x=-\dfrac{7}{2}\)
\(x=-\dfrac{7}{2}:2\)
\(x=-\dfrac{7}{2}.\dfrac{1}{2}\)
\(x=-\dfrac{7}{4}\)
b, \(\left(2x-3\right)^2=36\)
\(\left(2x-3\right)^2=6^2\)
\(2x-3=6\)
\(2x=9\)
\(x=\dfrac{9}{2}\)
c, \(5^{x+2}=625\)
\(5^{x+2}=5^4\)
\(x+2=4\)
\(x=2\)
Bài 1:
\(\Leftrightarrow n^2-1+2⋮n+1\)
\(\Leftrightarrow n+1\in\left\{1;2\right\}\)
hay \(n\in\left\{0;1\right\}\)
Dạng 1:
a: =>x(x-3)=0
=>x=3 hoặc x=0
b: =>x(3x-4)=0
=>x=4/3 hoặc x=0
c: =>2x-1=0
=>x=1/2
d: =>2x(2x+3)=0
=>x=0 hoặc x=-3/2
e: =>x(2x+5)=0
=>x=-5/2 hoặc x=0
\(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}\Rightarrow\)\(\frac{a^2}{4}=\frac{3b^2}{27}=\frac{2c^2}{32}\)
Áp dụng tc của dãy tỉ số bằng nhau ta có:
\(\frac{a^2}{4}=\frac{3b^2}{27}=\frac{2c^2}{32}=\frac{a^2+3b^2-2c^2}{4+27-32}=\frac{-16}{-1}=16\)
=> \(\frac{a^2}{4}=16\Rightarrow a=8\)
\(\frac{3b^2}{27}=16\Rightarrow b=12\)
\(\frac{2c^2}{32}=16\Rightarrow c=16\)
e) \(\frac{1}{7}.\frac{-3}{8}+\frac{-13}{8}.\frac{1}{7}\)
\(=\frac{1}{7}.\left[\left(-\frac{3}{8}\right)+\left(-\frac{13}{8}\right)\right]\)
\(=\frac{1}{7}.\left(-2\right)\)
\(=-\frac{2}{7}.\)
Chúc bạn học tốt!