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\(a,x^{10}=1\Leftrightarrow x=1\)
b, 2x = 256 <=> 2x = 28 <=> x = 8
c, x10 = x
<=> \(x^{10}-x=0\)
<=> \(x\left[x^9-1\right]=0\)
<=> x = 0 hoặc x = 1
d, \((2x-15)^5=(2x-15)^3\)
<=> \((2x-15)^5-(2x-15)^3=0\)
<=> \((2x-15)^2.\left[1-(2x-15)^3\right]=0\)
<=> \(\orbr{\begin{cases}2x-15=0\\1-(2x-15)^3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{15}{2}\\2x-15=\pm1\end{cases}}\)
Tìm nốt x đi .
Lâu lâu chưa dạng gặp dạng này
e) \(\frac{11.3^{22}.9.35-9.15}{\left(2.3^{14}\right)^2}\)
\(=\frac{11.3^{22}.3^2.5.7-3^2.3.5}{2^2.3^{28}}\)
\(=\frac{3^3.5.\left(11.3^{20}.7-1\right)}{2^2.3^{28}}\)
\(=\frac{5.\left(11.3^{20}.7-1\right)}{2^2.3^{25}}\)
Đề bài sai ko vậy ?? kết quả ko có ra phân số hoặc số nguyên mà là số mà bạn chưa học đâu
5^6+5^7+5^8
=5^6.(1+5+5^2)
=5^6.31 chia hết cho 31
7^6+7^5-7^4
=7^4.(7^2+7-1)
=7^4.55 chia hết cho 11
BÀI 2:
a) \(5^6+5^7+5^8=5^6\left(1+5+5^2\right)=5^6.31\) \(⋮\)\(31\)
b) \(7^6+7^5-7^4=7^4.\left(7^2+7-1\right)=7^4.55\)\(⋮\)\(11\)
c) \(2^3+2^4+2^5=2^3.\left(1+2+2^2\right)=2^3.7\)\(⋮\)\(7\)
d) mk chỉnh đề
\(1+2+2^2+2^3+...+2^{59}\)
\(=\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^{58}+2^{59}\right)\)
\(=\left(1+2\right)+2^2\left(1+2\right)+...+2^{58}\left(1+2\right)\)
\(=\left(1+2\right)\left(1+2^2+...+2^{58}\right)\)
\(=3\left(1+2^2+...+2^{58}\right)\)\(⋮\)\(3\)
a, \(M=1+6+6^2+6^3+...+6^{99}\)
\(M=6\cdot(1+6)+6^2(1+6)+6^3(1+6)+...+6^{99}(1+6)\)
\(M=6\cdot7+6^2\cdot7+6^3\cdot7+...+6^{99}\cdot7\)
\(M=7\cdot\left[6+6^2+6^3+...+6^{99}\right]⋮7(đpcm)\)
b, \(M=1+6+6^2+6^3+...+6^{99}\)
\(M=6\cdot\left[1+6+6^2+6^3\right]+...+6^{96}\left[1+6+6^2+6^3\right]\)
\(M=6\cdot\left[7+36+216\right]+...+6^{96}\left[7+36+216\right]\)
\(M=6\cdot259+...+6^{96}\cdot259\)
\(M=259\cdot\left[6+...+6^{96}\right]⋮259\)
Vậy \(M⋮259(đpcm)\)
Câu a:
5.2\(^2\) + x + 3 = 5\(^2\)
5.4 + x +3 = 25
20 + x + 3 = 25
x = 25- 20 - 3
x = 5 - 3
x = 2
Vậy x = 2
Câu b:
2\(^3\) + x - 3\(^2\) = 5\(^3\) - 4\(^3\)
8 + x - 9 = 125 - 64
x = 125 - 64 + 9 - 8
x = 61 + 9 - 8
x = 70 - 8
x = 62
Vậy x = 62
a) \(63^7< 64^7=\left(2^6\right)^7=2^{42}< 2^{48}=\left(2^4\right)^{12}=16^{12}\Rightarrow63^7< 16^{12}\)
b) \(3^{151}>3^{150}=\left(3^2\right)^{75}=9^{75}>8^{75}=\left(2^3\right)^{75}=2^{225}\)
c) \(9^{20}=\left(3^2\right)^{20}=3^{40}>3^{39}=\left(3^3\right)^{13}=27^{13}\Rightarrow9^{20}>27^{13}\)
bài 2:
a)\(2^x=32\Leftrightarrow2^x=2^5\Leftrightarrow x=5\)
b)\(2x+3^4=7^2\Leftrightarrow2x+81=49\Leftrightarrow2x=-32\Leftrightarrow x=-16\)
c)\(12x-33=3^2\Leftrightarrow12x-33=9\Leftrightarrow12x=42\Leftrightarrow x=\frac{7}{2}\)
a, T=312.46/610
T=312.(22)6/(2.3)10
T=312.212/210.310
T=32.22
T=9.4
T=36
Câu a:
T = 3\(^{12}\). \(\frac{4^6}{6^{10}}\)
T = \(\frac{3^{12.}2^{12}}{3^{10}.2^{10}}\)
T = 3^2.2^2
T = 9.4
T = 36
Bài 3:
a: \(A=4x^2+4x+11\)
\(=4x^2+4x+1+10\)
\(=\left(2x+1\right)^2+10\ge10\forall x\)
Dấu '=' xảy ra khi 2x+1=0
=>2x=-1
=>\(x=-\frac12\)
b: \(B=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
\(=\left(x^2+5x\right)^2-36\ge-36\forall x\)
Dấu '=' xảy ra khi \(x^2+5x=0\)
=>x(x+5)=0
=>x=0 hoặc x=-5
c: \(C=x^2-2x+y^2-4y+7\)
\(=x^2-2x+1+y^2-4y+4+2\)
\(=\left(x-1\right)^2+\left(y-2\right)^2+2\ge2\forall x,y\)
Dấu '=' xảy ra khi x-1=0 và y-2=0
=>x=1 và y=2
Bài 4:
a: \(A=5-8x-x^2\)
\(=-x^2-8x-16+21\)
\(=-\left(x+4\right)^2+21\le21\forall x\)
Dấu '=' xảy ra khi x+4=0
=>x=-4
b: \(B=5-x^2+2x-4y^2-4y\)
\(=-x^2+2x-1-4y^2-4y-1+7\)
\(=-\left(x-1\right)^2-\left(2y+1\right)^2+7\le7\forall x,y\)
Dấu '=' xảy ra khi x-1=0 và 2y+1=0
=>x=1 và y=-1/2
Bài 5:
a: \(a^2+b^2+c^2=ab+ac+bc\)
=>\(2\left(a^2+b^2+c^2\right)=2\left(ab+ac+bc\right)\)
=>\(2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)
=>\(\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)=0\)
=>\(\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\)
=>a=b=c
b: \(a^2-2a+b^2+4b+4c^2-4c+6=0\)
=>\(a^2-2a+1+b^2+4b+4+4c^2-4c+1=0\)
=>\(\left(a-1\right)^2+\left(b+2\right)^2+\left(2c-1\right)^2=0\)
=>a-1=0 và b+2=0 và 2c-1=0
=>a=1 và b=-2 và c=1/2
Bài 1:
a: \(A=100^2-99^2+98^2-97^2+\cdots+2^2-1^2\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+\cdots+\left(2-1\right)\left(2+1\right)\)
=100+99+98+87+...+2+1
\(=100\cdot\frac{\left(100+1\right)}{2}=5050\)
b: \(B=3\left(2^2+1\right)\left(2^4+1\right)\cdot\ldots\cdot\left(2^{64}+1\right)+1\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\cdot\ldots\cdot\left(2^{64}+1\right)+1\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)
\(=\left(2^{32}-1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)
\(=\left(2^{64}-1\right)\left(2^{64}+1\right)+1=2^{128}-1+1=2^{128}\)
c: \(C=\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\)
\(=\left(a+b\right)^2+2c\left(a+b\right)+c^2+\left(a+b\right)^2-2c\left(a+b\right)+c^2-2\left(a+b\right)^2\)
\(=2c^2\)
Bài 2:
a: \(\left(a+b\right)^3-3ab\left(a+b\right)\)
\(=a^3+3a^2b+3ab^2+b^3-3ab^2-3a^2b\)
\(=a^3+b^3\)
b: \(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left(a+b+c\right)\left\lbrack\left(a+b\right)^2-c\left(a+b\right)+c^2\right\rbrack-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
=(a+b+c)\(\left(a^2+b^2+c^2-ab-ac-bc\right)\)