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a) 4.(-5)2+(-2)3.25
= 4.25+(-8).25
=25.[4+(-8)]
=25.(-4)
=-100
b)\(15\dfrac{3}{7}-\left(\dfrac{7}{15}+9\dfrac{4}{7}\right)\)
= \(15\dfrac{3}{7}-\dfrac{7}{15}-9\dfrac{4}{7}\)
= \(\left(15\dfrac{3}{7}-9\dfrac{4}{7}\right)-\dfrac{7}{15}\)
=\(\left(14\dfrac{10}{7}-9\dfrac{4}{7}\right)-\dfrac{7}{15}\)
=\(5\dfrac{1}{7}-\dfrac{7}{15}\)
=\(\dfrac{36}{7}-\dfrac{7}{15}\)
=\(\dfrac{540}{105}-\dfrac{49}{105}\)
=\(\dfrac{491}{105}\)
\(\)a) 4.(-5)2+(-2)3.25
\(=4.5^2+\left(-2\right)^3.25\)
\(=4.25+\left(-8\right).25\)
\(=100+\left(-200\right)\)
\(=-100\)
b) \(15\dfrac{3}{7}-\left(\dfrac{7}{15}+9\dfrac{4}{7}\right)\)
\(=\dfrac{108}{7}-\left(\dfrac{7}{15}+\dfrac{67}{7}\right)\)
\(=\dfrac{108}{7}-\dfrac{1054}{105}\)
\(=\dfrac{566}{105}\)
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
Câu 2:
a: \(\Leftrightarrow12x-60=7x-5\)
=>5x=55
=>x=11
b: \(\Leftrightarrow\left(2x-3\right)^{2010}\left[\left(2x-3\right)^2-1\right]=0\)
=>(2x-3)(2x-2)(2x-4)=0
hay \(x\in\left\{\dfrac{3}{2};1;2\right\}\)
a,(3/5+0,415-3/200).\(2\dfrac{2}{3}\).0,25
=(0,6+0.415-3/200)8/3.1/4
=(1,015-3/200).8/3.1/4
=(1015/1000-3/200).2/3
=(203/200-3/200).2/3
=1.2/3=2/3
b,0,25:(10,3-9,8)-3/4
=0,25:910,3-9,8)-0.75
=0,25:0,5-0,75
=-0,25
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\(\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{59.61}\)
= \(\dfrac{2}{2}.\left(\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{59.61}\right)\)
= \(\dfrac{3}{2}.\left(\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{59.61}\right)\)
= \(\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\)
= \(\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{61}\right)\)
=\(\dfrac{3}{2}.\dfrac{56}{305}\)
= \(\dfrac{78}{305}\)
\(\left(x^2-4\right)\left(6-2x\right)=0\) ⇔ \(x^2-4=0\) 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 }
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
Câu 2:
a: \(3\dfrac{1}{3}x+16=13,25\)
\(\Leftrightarrow x\cdot\dfrac{10}{3}=-3,25\)
hay \(x=-\dfrac{13}{4}:\dfrac{10}{3}=\dfrac{-39}{40}\)
b: \(\left(2x+\dfrac{1}{3}\right)^3=-\dfrac{8}{27}\)
=>2x+1/3=-2/3
=>2x=-1
hay x=-1/2