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a)38:3^4+2^2.2^3=3^8-4+2^2+3=3^4+2^5=81+32=113
b)3.4^2-2.3^2=3(4^2-2.3)=3(16-6)=3.10=30
c)4^6.3^4.9^5/6^12=(2^2)^6.3^4.(3^2)^5/(2.3)^12=2^12.3^4.3^10/2^12.3^12=2^12.3^14/2^12.3^12=3^14/3^12=3^2=9
e)45^3.20^4.18^2/180^5=(5.3^2)^3.(2^2.5)^4.(2.3^2)^2/(2^2.3^2.5)^5=5^3.3^6.2^8.5^4.2^2.3^4/2^10.3^10.5^5=5^9.3^10.2^10/2^10.3^10.5^5
=5^4/1=5^4=625
d)21^2.14.125/35^3.6=(2.7)^2.2.7.5^3/(5.7)^3.2.3=2^2.7^2.2.7.5^3/5^3.7^3.2.3=2^3.7^3.5^3/5^3.7^3.2.3=2^3/2.3=8/6=4/3
g)2^13+2^5/2^10+2^2=2^5(2^8+1)/2^2(2^8+1)=2^5/2^2=32/4=8
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Bài 1:
a: =>13x+8=9x+20
=>4x=12
hay x=3
b: \(\Leftrightarrow5x-7=-8-11-3x\)
=>5x-7=-3x-19
=>8x=-12
hay x=-3/2
c: \(\Leftrightarrow\left[{}\begin{matrix}12x-7=5\\12x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{6}\end{matrix}\right.\)
e: =>3x+1=-5
=>3x=-6
hay x=-2
a: =>5x=3x-6
=>2x=-6
hay x=-3
b: \(\Leftrightarrow\left(x-3\right)^2=4\cdot5^2=100\)
=>x-3=10 hoặc x-3=-10
=>x=13 hoặc x=-7
c: \(\left|x^3+1\right|+2\ge2\forall x\)
Dấu '=' xảy ra khi x=-1
\(\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{59.61}\)
= \(\dfrac{2}{2}.\left(\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{59.61}\right)\)
= \(\dfrac{3}{2}.\left(\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{59.61}\right)\)
= \(\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\)
= \(\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{61}\right)\)
=\(\dfrac{3}{2}.\dfrac{56}{305}\)
= \(\dfrac{78}{305}\)
\(\left(x^2-4\right)\left(6-2x\right)=0\) ⇔ \(x^2-4=0\) hoặc \(6-2x=0\)
*Nếu \(x^2-4=0\)
⇒ x2 = 4
⇒ x ∈ {2 ; -2}
*Nếu \(6-2x=0\)
⇒2x = 6
⇒ x = 6 : 2 = 3
Vậy x ∈ { -2 ; 2 ; 3 }
a: \(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5+3^5}\cdot\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5+2^5+2^5+2^5+2^5}=2^x\)
\(\Leftrightarrow2^x=\dfrac{4^5}{3^5}\cdot\dfrac{6^5}{2^5}=4^5=2^{10}\)
=>x=10
b: \(\left(x-1\right)^{x+4}=\left(x-1\right)^{x+2}\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow x\left(x-1\right)^{x+2}\cdot\left(x-2\right)=0\)
hay \(x\in\left\{0;1;2\right\}\)
c: \(6\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\)
\(\Leftrightarrow5\cdot\left(6-x\right)^{2003}=0\)
\(\Leftrightarrow6-x=0\)
hay x=6
1.Tính hợp lý:
a. 1152 - (374 + 1152) + (374 - 65) = 1152 - 374 - 1152 + 374 - 65 = ( 1152 - 1152 ) + ( -65) + ( 374 - 374 ) = 0 + ( - 65) + 0 = -65
Bài 1 : Tính hợp lý : c. \(\dfrac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\) = \(\dfrac{11.3^{29}-3^{30}}{2^2.3^{28}}\) = \(\dfrac{3^{29}.\left(11-3\right)}{2^2.3^{28}}\) = \(\dfrac{3^{29}.2^3}{2^2.3^{28}}\) = 6
a, =\(3^4+2^5=81+32=113\)
b, =\(3.\left(4^2-2.3\right)=3.\left(16-6\right)=3.10=30\)
c, =\(\dfrac{2^{12}.3^4.3^{10}}{2^{12}.3^{12}}=\dfrac{2^{12}.3^{14}}{2^{12}.3^{12}}=3^2=9\)
d, =\(\dfrac{3^2.7^2.2.7.5^3}{5^3.7^3.2.3}=3\)
e, =\(\dfrac{3^6.5^3.2^8.5^4.2^2.3^4}{2^{10}.3^{10}.5^5}=\dfrac{3^{10}.2^{10}.5^7}{2^{10}.3^{10}.5^5}=5^2=25\)
g, =\(\dfrac{2^5.\left(2^8+1\right)}{2^2.\left(2^8+1\right)}=\dfrac{2^5}{2^2}=2^3=8\)
thank