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=(1+2+3+4+...+99+100)^2 =((100+1).100:2)^2=25502500 tick nha pan
\(\Leftrightarrow3.\left(7x^2+1\right)=4.\left(8x^2-2\right)\)
\(\Leftrightarrow21x^2+3=32x^2-8\)
\(\Leftrightarrow21x^2+3-32x^2+8=0\)
\(\Leftrightarrow-11x^2+11=0\)
\(\Leftrightarrow-11\left(x^2-1\right)=0\)
\(\Leftrightarrow x^2-1=0\)
\(\Leftrightarrow x^2=1\)
\(\Leftrightarrow x=1\)
#quankun^^
\(\left(7x^2+1\right):4=\left(8x^2-2\right):3\)
\(\frac{7x^2+1}{4}=\frac{8x^2-2}{3}\)
\(\left(7x^2+1\right).3=\left(8x^2-2\right).4\)
\(21x^2+3=32x^2-2\)
\(21x^2-32x^2=-2-3\)
\(-11x^2=-5\)
\(x^2=\frac{5}{11}\)
\(x^2=\sqrt{\frac{5}{11}}=\frac{\sqrt{55}}{11}\)
2x+1.3y=36x=(4.9)x=4x.9x=(22)x.(32)x=22x.32x
=>2x+1=22x=>x+1=2x=>2x-x=1=>x=1
và 3y=32x=>y=2x=>y=2.1=>y=2
Vậy (x;y)=(1;2)
ai giúp mik câu này vs ak
mik tick cho
M-(xy3-2xy+x2+5)=xy3+5xy-2x2-6
(xy3-2xy+x2+5)-M=xy3+5xy-2x2-6
1) = (2m)2 - 32 = (2m+3)(2m-3)
2) = x2 - (2y)2 = (x+2y)(x-2y)
Câu 1:
a) 2225 và 3150
Ta có:2225=(29)25=51225
3150=(36)25=72925
Vì 51225<72925
Suy ra: 2225<3150
Câu 2:
a)\(25^3:5^2=\left(5^2\right)^3:5^2=5^6:5^2=5^4\)
b)\(\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6=\left(\frac{3}{7}\right)^{21}:\left[\left(\frac{3}{7}\right)^2\right]^6=\left(\frac{3}{7}\right)^{21}:\left(\frac{3}{7}\right)^{12}=\left(\frac{3}{7}\right)^9\)
c)\(3-\left(-\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^2:2=3+\frac{1}{4}:2=3+\frac{1}{8}=\frac{25}{8}\)
Câu 3:
a)\(9.3^3.\frac{1}{81}.3^2=3^2.3^3.3^2.\left(\frac{1}{3^4}\right)=3^7:3^4=3^3\)
b)\(4.2^5:\left(2^3.\frac{1}{16}\right)=2^2.2^5:\left(2^3.\frac{1}{2^4}\right)=2^7:\frac{1}{2}=2^8\)
c)\(3^2.2^5.\left(\frac{2}{3}\right)^2=288.\frac{4}{9}=2^7\)
d)\(\left(\frac{1}{3}\right)^3.\frac{1}{3}.9^2=\left(\frac{1}{3}\right)^4.\left(3^2\right)^2=3^4.\left(\frac{1}{3}\right)^4=3^4:3^4=1\)