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Tìm x, biết
a/ ( x - 15 ) - 75 = 0
( x - 15 ) = 0 + 75
( x - 15 ) = 75
x = 75 + 15
x = 100
b/ 315 + ( 125 - x ) = 435
( 125 - x ) = 435 - 315
( 125 - x ) = 120
x = 125 - 120
x = 5
c/ 2031 - x + 102 = 1003
2031 - x = 1003 -102
2031 - x = 901
x = 2031 - 901
x = 1130
d/ x + 1002 = 1003 + 4327
x + 1002 = 5330
x = 5330 - 1002
x = 4328
Hok tốt
Trả lời :
a) (x - 15) - 75 = 0
x - 15 = 0 + 75
x - 15 = 75
x = 75 + 15
x = 90
(các câu kia tương tự)
1,
a,1100+(-100)=1000
b,(2017)+2010=-7
c,/-102/+36=138
d,/-1002/+(-102)=900
e,(-1002)+(-102)+515=589
a, BCNN(210;280) = 840 => BC(210;280) = 840k (k ∈ N)
b, BCNN(105;135) = 945 => BC(105;135) = 945k (k ∈ N)
c, BCNN(24;36) = 72 => BC(24;36) = 72k (k ∈ N)
d, BCNN(48;80;72) = 720 => BC(48;80;72) = 720k (k ∈ N)
e, BCNN(42;55;91) = 30030 => BC(42;55;91) = 30030k (k ∈ N)
f, BCNN(12;36;102) = 612 => BC(12;36;102) = 612k (k ∈ N)
a) 18 2 < 10 3
b) 3 2 + 4 2 < ( 3 + 4 ) 2
c) 100 2 + 30 2 < ( 100 + 30 ) 2
d) a 2 + b 2 > ( a - b ) 2 với a ∈ N * ; b ∈ N * .
Ta có:
\(x^2+y^2=1\Rightarrow\left(x^2+y^2\right)^2=1\)(1)
Thay (1) vào \(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\)ta có:
\(\frac{x^4}{a}+\frac{y^4}{b}=\frac{\left(x^2+y^2\right)^2}{a+b}\Leftrightarrow\frac{x^4b+y^4a}{ab}=\frac{x^4+2x^2y^2+y^4}{a+b}\)
\(\Leftrightarrow\left(x^4b+y^4a\right)\left(a+b\right)=\left(x^4+2x^2y^2+y^4\right).ab\)
\(\Leftrightarrow x^4ab+x^4b^2+y^4a^2+y^4ab=x^4ab+2x^2y^2ab+y^4ab\)
\(\Leftrightarrow x^4b^2+y^4a^2=2x^2y^2ab\)
\(\Leftrightarrow\left(x^2b\right)^2-2x^2y^2ab+\left(y^2a\right)^2=0\)
\(\Leftrightarrow\left(x^2b-y^2a\right)^2=0\)
\(\Leftrightarrow x^2b-y^2a=0\)
\(\Leftrightarrow x^2b=y^2a\)
\(\Rightarrow\frac{x^2}{a}=\frac{y^2}{b}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\)
\(\Rightarrow\left(\frac{x^2}{a}\right)^{1002}=\left(\frac{y^2}{b}\right)^{1002}=\left(\frac{1}{a+b}\right)^{1002}\)
\(\Rightarrow\frac{x^{2004}}{a^{1002}}=\frac{y^{2004}}{b^{1002}}=\frac{1}{\left(a+b\right)^{1002}}\)
\(\Rightarrow\frac{x^{2004}}{a^{1002}}+\frac{y^{2004}}{b^{1002}}=\frac{1}{\left(a+b\right)^{1002}}+\frac{1}{\left(a+b\right)^{1002}}=\frac{2}{\left(a+b\right)^{1002}}\left(đpcm\right)\)
Chúc bạn học tốt!
a)311x167+689x293+126x311
=311x(167 + 126)+689x293
=311x293+689x293
=(311+689)x293
=1000x293
=293000
b)44x55+147x156-184x78
=44x55+147x156-92x2x78
=44x55+147x156-92x156
=44x55+(147-92)x156
=44x55+55x156
=55x(44+156)
=55x200
=11000
c)125x11x56-286x77-7x154
=125x11x56-286x77-7x11x14
=125x11x56-286x77-77x14
=125x11x56-(286-14)x77
=125x11x56-272x77
=125x11x7x8-272x77
=125x77x8-272x77
=1000x77-272x77
=(1000-272)x77
=738x77
=56056
d)(42+72+102+...+1002)x(1212x27-2727x12)
Đặt A là (42+72+102+...+1002), B là (1212x27-2727x12)
Với A:42+72+102+...+1002
=42+42x32+42x62+...+42x962
=42x(32+62+...+962)
=42x(9+36+...+9216)
=42x95391
=16x95391
=1526256
Với B:1212x27-2727x12
=101x12x27-101x27x12
=0
Vậy AxB=1526256x0=0