12 : 5 ; 75 : 12 ; 23 : 4 ; 81 : 4
5+5+5+5+5+5+5+5+5-6-6-6-6-6=
12+12+12+12+12+12+12+12+12+12=
5+5+5+5+5+5+5+5+5-6-6-6-6-6=5*9-[6*5]
=45-30
=15
12+12+12+12+12+12+12+12+12+12=12*10
=120
đóng ngoặc 6*5 để làm j vậy bn
???
???
=5 x 9 = 45
=6 x 5 = 30
45 - 30 = 15
12 x 10 = 120
(546+545+544):(539.55)
(1259+1258+1257):(1256+1257+1258)
\(\left(5^{46}+5^{45}+5^{44}\right):\left(5^{39}.5^5\right)=5^{44}.\left(1+5+5^2\right):5^{44}=1+5+5^2=31\)
\(\left(12^{59}+12^{58}+12^{57}\right):\left(12^{56}+12^{57}+12^{58}\right)\)
=\(12^{57}.\left(12^2+12+1\right):\left[12^{56}.\left(1+12+12^2\right)\right]\)
=12
**** mình nha bạn !!!!!!!!!
\( $\frac{3^{14}.5^{10}-3^3}{3^{12}.5^{12}+5^{12}.3^{ }^{12}^{ }}=\frac{3^{11}.5^{10}-1}{2.3^9.5^{12}}$\)
a)(546+545+544):(539.55)
b)(1259+1258+1257):(1256+1257+1258)
Bài 1 : tính
a)\(\sqrt{49-12\sqrt{5}}+\sqrt{49+12\sqrt{5}}\)
b)\(\sqrt{29+12\sqrt{5}}+\sqrt{29-12\sqrt{5}}\)
a) \(=\sqrt{\left(3\sqrt{5}-2\right)^2}+\sqrt{\left(3\sqrt{5}+2\right)^2}=3\sqrt{5}-2+3\sqrt{5}+2=6\sqrt{5}\)
b) \(=\sqrt{\left(2\sqrt{5}+3\right)^2}+\sqrt{\left(2\sqrt{5}-3\right)^2}=2\sqrt{5}+3+2\sqrt{5}-3=4\sqrt{5}\)
tính D bt D= [(-5)+(-12)] +[(-12)+5] +[-5+12]
D= [(-5)+(-12)] +[(-12)+5] +[-5+12]
D= -17+ (-7)+ 7
D= -17
CHÚC BẠN HOK TỐT
Mọi người check xem đúng không:
<=>x(3x+2)+(x+1)^2 - (2x - 5)(2x+5)= -12
<=>3x^2 + 2x + (x+1)^2 - 2x^2 - 5^2= -12
<=>(3x^2 - 2x^2) + [ (x +1)^2 -5^2] + 2x = -12
<=>(3x - 2x)(3x + 2x)+ (x+1-5) (x+1+5) + 2x = -12
=> 3x - 2x = -12 ; 3x+2x = -12 ; x+1+5 = -12 ; x+1-5 = -12 hoặc 2x = -12
=> x = -12 ; 5x =-12 ; x+ 6 = -12 ; x -4 = -12 hoặc x = -12: 2
=> x= -12 ; x = -12:5; x = -12 :6; x = -12 + 4 hoặc x= -6
=> x= -12 x = -12/5; x = -2 ; x = -8 hoặc x = -6
đúng rồi nha bạn
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Mọi người check xem đúng không:
<=>x(3x+2)+(x+1)^2 - (2x - 5)(2x+5)= -12
<=>3x^2 + 2x + (x+1)^2 - 2x^2 - 5^2= -12
<=>(3x^2 - 2x^2) + [ (x +1)^2 -5^2] + 2x = -12
<=>(3x - 2x)(3x + 2x)+ (x+1-5) (x+1+5) + 2x = -12
=> 3x - 2x = -12 ; 3x+2x = -12 ; x+1+5 = -12 ; x+1-5 = -12 hoặc 2x = -12
=> x = -12 ; 5x =-12 ; x+ 6 = -12 ; x -4 = -12 hoặc x = -12: 2
=> x= -12 ; x = -12:5; x = -12 :6; x = -12 + 4 hoặc x= -6
=> x= -12 x = -12/5; x = -2 ; x = -8 hoặc x = -6
bạn làm sai rồi !
\(\Leftrightarrow x\left(3x+2\right)+\left(x+1\right)^2-\left(2x-5\right)\left(2x+5\right)=-12\)
\(\Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25=-12\)
\(\Leftrightarrow4x+26=-12\)
\(\Leftrightarrow4x=-38\)
\(\Leftrightarrow x=-\frac{19}{2}\)
Vậy tập nghiệm của phương trình là \(S=\left\{-\frac{19}{2}\right\}\)
Tính (theo mẫu).
Mẫu: \(\dfrac{1}{2}-\dfrac{5}{12}=\dfrac{6}{12}-\dfrac{5}{12}=\dfrac{6-5}{12}=\dfrac{1}{12}\) |
a) \(\dfrac{3}{4}-\dfrac{1}{8}\) b) \(\dfrac{2}{6}-\dfrac{5}{18}\) c) \(\dfrac{2}{5}-\dfrac{3}{20}\)
a) \(\dfrac{3}{4}-\dfrac{1}{8}=\dfrac{6}{8}-\dfrac{1}{8}=\dfrac{6-1}{8}=\dfrac{5}{8}\)
b) \(\dfrac{2}{6}-\dfrac{5}{18}=\dfrac{6}{18}-\dfrac{5}{18}=\dfrac{6-5}{18}=\dfrac{1}{18}\)
c) \(\dfrac{2}{5}-\dfrac{3}{20}=\dfrac{8}{20}-\dfrac{3}{20}=\dfrac{8-3}{20}=\dfrac{5}{20}=\dfrac{1}{4}\)
\(\dfrac{-5}{12}\).\(\dfrac{7}{15}\)\(+\)\(\dfrac{-5}{12}\)\(+\)\(\dfrac{8}{15}\)\(+\)\(\dfrac{5}{12}\)
\(\dfrac{-5}{12}\cdot\dfrac{7}{15}+\dfrac{-5}{12}\cdot\dfrac{8}{15}+\dfrac{5}{12}\)
\(=\dfrac{-5}{12}+\dfrac{5}{12}\)
=0