x mũ 2+(x-2)mũ 2=10 mũ 2
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a) \(\dfrac{10^{12}+5^{11}.2^9-5^{13}.2^8}{4.5^5.10^6}\)
\(=\dfrac{2^{12}.5^{12}+5^{11}.2^9-5^{13}.2^8}{2^2.5^5.2^6.5^6}\)
\(=\dfrac{2^{12}.5^{12}+5^{11}.2^9-5^{13}.2^8}{2^8.5^{11}}\)
\(=\dfrac{\left(2^8.5^{11}\right)\left(2^4.5+2-5^2\right)}{2^8.5^{11}}\)
\(=2^4.5+2-5^2\)
\(=57\)
b) \(\dfrac{\left[5\left(x-y\right)^4-3\left(x-y\right)^3+4\left(x-y\right)^2\right]}{\left(y-x\right)^2}\)
\(=\dfrac{\left(x-y\right)^2\left[5\left(x-y\right)^2-3\left(x-y\right)+4\right]}{\left(y-x\right)^2}\)
\(=\dfrac{\left(x^2+y^2-2xy\right)\left[5\left(x-y\right)^2-3\left(x-y\right)+4\right]}{\left(y^2+x^2-2xy\right)}\)
\(=5\left(x-y\right)^2-3\left(x-y\right)+4\)
c) \(\dfrac{\left(x+y\right)^5-2\left(x+y\right)^4+3\left(x+y\right)^3}{-5\left(x+y\right)^3}\)
\(=\dfrac{\left(x+y\right)^3\left[5\left(x+y\right)^2-2\left(x+y\right)+3\right]}{-5\left(x+y\right)^3}\)
\(=\dfrac{5\left(x+y\right)^2-2\left(x+y\right)+3}{-5}\)
Bài 1:
2\(x\) = 4
2\(^x\) = 22
\(x=2\)
Vậy \(x=2\)
Bài 2:
2\(^x\) = 8
2\(^x\) = 23
\(x=3\)
Vậy \(x=3\)
\(2^3\cdot2^2\cdot2^x\cdot x^5\cdot=2^{5+x}\cdot x^5\)
\(10^2\cdot2^{10}\cdot10^3\cdot10^5=10^{10}\cdot2^{10}=2^{10}\cdot5^{10}\cdot2^{10}=4^{10}\cdot5^{10}=20^{10}\)
\(a^3\cdot a^2\cdot a^5=a^{3+2+5}=a^{10}\)
P/s: Mình chỉ hiểu ý bạn như này!
\(\left\{10^2-\left[3^2.5+2^3.\left(4+2^{12}:2^{11}\right)\right]\right\}.2\)
\(=\left\{100-\left[3^2.5+2^3.\left(4+2\right)\right]\right\}.2\)
\(=\left\{100-\left[9.5+8.6\right]\right\}.2\)
\(=\left\{100-93\right\}.2\)
\(=7.2=14\)
\(\left\{10^2-\left[3^2\times5+2^3\times\left(4+2^{12}:2^{11}\right)\right]\right\}\times2\)
\(=\left\{100-\left[9\times5+8\times\left(4+2\right)\right]\right\}\times2\)
\(=\left\{100-\left[9\times5+8\times6\right]\right\}\times2\)
\(=\left\{100-93\right\}\times2\)
\(=7\times2\)
\(=14\)
1) 62 . x = 36
=> x = 36 / 36 = 1
2) 9x - 42 = 11
9x - 16 = 11
9x = 11 + 16 =27
=> x = 27/9=3
3) 10 + 2x = 42
2x = 16 - 10 =6
=> x = 3
4) 231 - ( x - 6 ) =103
( x - 6 ) = 231-103=128
=> x = 128+6=134
5) 10 + 3x = 45 / 42
3x = 43-10=64-10=54
x = 18
6)25 + 52x = 82 + 62
25 + 25x = 100
25x = 100-25=75
=> x = 75/25=3
*Chúc bạn học tốt*
1) 62.x=36 2) 9.x-42=11 3)10+2.x=42 4)231-(x-6)=103 5)10+3x=45:42 6)25+52.x=82+62
36.x=36 9.x-16=11 10+2.x=16 x-6 = 231-103 10+3x=43 25+25.x =64+36
x=36:36 9.x =11+16 2.x =16-10 x-6 =128 10+3x=64 25+25.x =100
x=1 9.x =27 2.x =6 x =128+6 3x=64-10 25.x =100-25
x =27:9 x =6:2 x =134 3x=54 25.x =75
x =3 x =3 x=54:3 x =75:25
x=18 x =3
Câu a mình ko bt trình bày thông cảm
b) \(^{2^x.\left(2^2\right)^2=\left(2^3\right)^2}\)
\(2^x.2^4=2^6\)
\(2^x=2^6:2^4\)
\(2^x=2^2\)
\(x=2\)
\(x^2+\left(x-2\right)^2=10^2\)
\(\Leftrightarrow x^2+x^2-2\cdot2\cdot x+2^2-10^2=0\)
\(\Leftrightarrow2x^2-4x-96=0\)
\(\Delta'=b'^2-ac=\left(-2\right)^2-2\cdot\left(-96\right)=4+192=196\)
\(\Delta'>0\)nên phương trình đã cho có hai nghiệm phân biệt :
\(x_1=\frac{-b'+\sqrt{\Delta'}}{a}=\frac{2+\sqrt{196}}{2}=8\)
\(x_2=\frac{-b'-\sqrt{\Delta'}}{a}=\frac{2-\sqrt{196}}{2}=-6\)
Vậy \(S=\left\{8;-6\right\}\)
\(x^2+\left(x-2\right)^2=10^2\)
\(\Leftrightarrow x^2+\left(x-2\right)^2=100\)
\(\Leftrightarrow x^2+x^2-4x+4-100=0\)
\(\Leftrightarrow2x^2+4x-96=0\Leftrightarrow2\left(x-6\right)\left(x+8\right)=0\)
TH1 : \(x-6=0\Leftrightarrow x=6\)
TH2 : \(x+8=0\Leftrightarrow x=-8\)