Tim x :
\(a.\left\{\frac{1}{2}-\frac{1}{3}\right\}.6^x+6^{x+2}=6^{10}+6^7\)
\(b.\left\{\frac{1}{2}-\frac{1}{6}\right\}.3^{x+4}-4.3^x=3^{16}-4.3^{13}\)
ai nhanh mk tk !
1.tìm số nguyên x biết:
a)\(\left(\frac{1}{2}-\frac{1}{3}\right).6^x+6^{x+2}=6^{15}+6^{18}\)
b)\(\left(\frac{1}{2}-\frac{1}{6}\right).3^{x+4}-4.3^x=3^{16}-4.3^{13}\)
giúp mk với
a)\(\left(\frac{1}{2}-\frac{1}{3}\right).6^x+6^{x+2}=6^{15}+6^{18}\)
\(\frac{1}{6}.6^x+6^{x+2}=6^{15}\left(1+6^3\right)\)
\(\frac{1}{6}.6^x\left(1+6^3\right)=6^{15}.217\)
\(6^{x-1}.217=6^{15}.217\)
\(6^{x-1}=6^{15}\)
\(x-1=15\)
\(x=16\)
b) \(\left(\frac{1}{2}-\frac{1}{6}\right).3^{x+4}-4.3^x=3^{16}-4.3^{13}\)
\(\frac{1}{3}.3^x.4\left(3^4-1\right)=3^{13}.4\left(3^3-1\right)\)
\(3^x.4.\left(3^3-1\right)=3^{13}.4.\left(3^3-1\right)\)
\(3^x=3^{13}\)
\(x=13\)
\(\left(\frac{1}{2}-\frac{1}{6}\right).\left(3^x.3^4\right)-4.3^x=3^{16}-4.3^{13}\)
=> \(\frac{1}{3}.3^x.3^4-4.3^x=3^{16}-4.3^{13}\)
=> \(3^x.3^4-4.3^x=\left(3^{16}-4.3^{13}\right):\frac{1}{3}\)
=> \(3^x.3^4-4.3^x=-386339074,3\)
=> \(3^x.\left(3^4-4\right)=-386339074,3\)
=> \(3^x.77=-386339074,3\)
=> \(3^x=-386339074,3:77\)
=> \(3^x=-5017390,575\)
=> x = ... chắc tự ngồi tính đc
Tìm số nguyên x, nếu biết
\(5^{x+4}-3.5^{x+3}=2.5^{11}\)
\(4^{x+3}-3.4^{x+1}=13.4^{11}\)
\(\left(\frac{1}{3}+\frac{1}{6}\right)2^{x+4}-2^x=2^{13}-2^{10}\)
\(\left(\frac{1}{2}-\frac{1}{6}\right)3^{x+4}-4.3^x=3^{16}-4.3^{13}\)
\(5^{x+4}-3.5^{x+3}=2.5^{11}\)
\(5^{x+3}\left(5-3\right)=2.5^{11}\)
\(5^{x+3}.2=2.5^{11}\)
\(5^{x+3}=5^{11}\)
\(x+3=11\)
\(x=8\)
\(4^{x+3}-3.4^{x+1}=13.4^{11}\)
\(4^{x+1}\left(4^2-3\right)=13.4^{11}\)
\(4^{x+1}.13=13.4^{11}\)
\(4^{x+1}=4^{11}\)
\(x+1=11\)
\(x=10\)
Tìm x biết :
\(a,2^x=\frac{8^4}{16^3}\)
\(b,\left(-2\right)^x=-4^6-8^5\)
\(c,4^x=4.4^{10}-3.4^{10}\)
\(d,4.3^{x-1}+2.3^{x+2}=4.3^6+2.3^9\)
\(e,\left(\frac{1}{3}+\frac{1}{6}\right).2^x+2^{x+1}=2^{12}+2^{10}\)
Ai nhanh mk sẽ tik nha , cám ơn m.n nhiều lắm
a, => 2^x = (2^3)^4/(2^4)^3 = 2^12/2^12 = 1 = 2^0
=> x = 0
c, => 4^x = 4^10.(4-3) = 4^10
=> x=10
d, => 2^2.3^x-1 + 2.3^x.9 = 2^2.3^6+2.3^9
=> 2.3^x-1 . (2+3.9) = 2.3^6.(2+3^3)
=> 2.3^x-1 . 27 = 2.3^6 . 27
=> 3^x-1 = 3^6
=> x-1 = 6
=> x = 7
e, => 2^x.(1/3+1/6+2) = 2^11.(2+1/2)
=> 2^x. 5/2 = 2^11. 5/2
=> 2^x = 2^11
=> x = 11
Tk mk nha
câu b) chưa có ai làm thì mình làm nốt vậy
\(\left(-2\right)^2=-4^6-8^5\)
\(\left(-2\right)^x=-4096-32768\)
\(\left(-2\right)^x=-36864\)
\(\Rightarrow x\) sẽ 1 số thập phân nào đó
\(a,\left(-3\right)^{x+3}=-\frac{1}{27}\)
\(b,\left(-6\right)^{2x+2}=\frac{1}{36}\)
\(c,\left(-3\right)^{x+5}=\frac{1}{81}\)
\(d,\left(\frac{1}{9}^x\right)=\left(\frac{1}{27}\right)^6\)
\(e,\left(\frac{4}{9}\right)^x=\left(\frac{8}{27}\right)^6\)
\(f,5^{x+4}-3.5^{x+3}=2.5^{11}\)
\(r,4.3^{x-1}+2.3^{x+2}=4.3^6+2.3^9\)
\(h,\left(\frac{1}{2}-\frac{1}{3}\right).6x+6^{x+2}=6^{10}+6^7\)
nhờ mấy bn giúp mk tối mình nộp rồi
a)\(\left(-3\right)^{x+3}=-\frac{1}{27}\)
\(\left(-3\right)^{x+3}=\left(-\frac{1}{3}\right)^3\)
\(\left(-3\right)^{x+3}=\left(-\frac{3^0}{3^1}\right)^3\)
\(\left(-3\right)^{x+3}=\left(-3^{-1}\right)^3\)
\(\left(-3\right)^{x+3}=\left(-3\right)^{-3}\)
\(\Rightarrow x+3=-3\)
\(\Rightarrow x=-6\)
b)\(\left(-6\right)^{2x+2}=\frac{1}{36}\)
\(\left(-6\right)^{2x+2}=\left(-\frac{1}{6}\right)^2\)
\(\left(-6\right)^{2x+2}=\left(-\frac{6^0}{6^1}\right)^2\)
\(\left(-6\right)^{2x+2}=\left(-6^{-1}\right)^2\)
\(\left(-6\right)^{2x+2}=\left(-6\right)^{-2}\)
\(\Rightarrow2x+2=-2\)
\(\Rightarrow2x=-4\)
\(\Rightarrow x=-2\)
c)\(\left(-3\right)^{x+5}=\frac{1}{81}\)
\(\left(-3\right)^{x+5}=\left(-\frac{1}{3}\right)^4\)
\(\left(-3\right)^{x+5}=\left(-\frac{3^0}{3^1}\right)^4\)
\(\left(-3\right)^{x+5}=\left(-3^{-1}\right)^4\)
\(\left(-3\right)^{x+5}=\left(-3\right)^{-4}\)
\(\Rightarrow x+5=-4\)
\(\Rightarrow x=-9\)
d)\(\left(\frac{1}{9}\right)^x=\left(\frac{1}{27}\right)^6\)
\(\left[\left(\frac{1}{3}\right)^2\right]^x=\left[\left(\frac{1}{3}\right)^3\right]^6\)
\(\left(\frac{1}{3}\right)^{2x}=\left(\frac{1}{3}\right)^{18}\)
\(\Rightarrow2x=18\)
\(\Rightarrow x=9\)
e)\(\left(\frac{4}{9}\right)^x=\left(\frac{8}{27}\right)^6\)
\(\left[\left(\frac{2}{3}\right)^2\right]^x=\left[\left(\frac{2}{3}\right)^3\right]^6\)
\(\left(\frac{2}{3}\right)^{2x}=\left(\frac{2}{3}\right)^{18}\)
\(\Rightarrow2x=18\)
\(\Rightarrow x=9\)
f)\(5^{x+4}-3\cdot5^{x+3}=2\cdot5^{11}\)
\(5^{x+3}\cdot5-3\cdot5^{x+3}=2\cdot5^{11}\)
\(5^{x+3}\left(5-3\right)=2\cdot5^{11}\)
\(5^{x+3}\cdot2=2\cdot5^{11}\)
\(\Rightarrow5^{x+3}=5^{11}\)
\(\Rightarrow x+3=11\)
\(\Rightarrow x=8\)
r)\(4\cdot3^{x-1}+2\cdot3^{x+2}=4\cdot3^6+2\cdot3^9\)
\(4\cdot3^x:3+2\cdot3^x\cdot9=4.3^7:3+2\cdot3^7\cdot9\)
\(3^x\left(4:3+2\cdot9\right)=3^7\left(4:3+2\cdot9\right)\)
\(\Rightarrow3^x=3^7\)
\(\Rightarrow x=7\)
1. tìm số nguyên x biết:
a)\(\left(\frac{1}{2}-\frac{1}{3}\right).6^x+6^{x+2}=6^{15}+6^{18}\)
b)\(\left(\frac{1}{2}-\frac{1}{6}\right).3^{x+4}-4.3^x=3^{16-4}.3^{13}\)
giúp mk nha
a.\(\frac{1}{6}.6^x+6^x.36=6^{15}\left(1+6^3\right)\)
\(6^x.\frac{217}{6}=6^{15}.217\)
\(6^x=6^{16}\)
\(x=16\)
a)A=\(\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}\) b)B=\(\frac{45}{19}-\left(\frac{1}{2}\left(\frac{1}{3}+\left(\frac{1}{4}\right)^{-1}\right)^-\right)^{-1}\) c)C=\(\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^{10}.6^{19}-7.2^{29}.27^6}\)
d)D=\(\frac{2^{21}.3^5-4^6.81}{\left(2^2.3\right)^6+8^4.3^5}\) e) E=\(\left(6^9.2^{10}+12^{10}\right):\left(2^{19}.27^3+15.4^9.9^4\right)\)
f) F=\(\frac{3^6.45^4-15^{13}.5^{-9}}{27^4.24^3+45^6}\) g)G=\(\frac{\left(\frac{2}{5}\right)^7.5^7+\left(\frac{9}{4}\right)^3:\left(\frac{3}{16}\right)^3}{2^7.5^2+512}\) h)H=\(x+\frac{0,2-0,375+\frac{5}{11}}{-0,3+\frac{9}{16}-\frac{15}{22}}\)với x=-1/3
ai nhanh nhất mà trả lời dúng mik tặng 3 k
Bài 1 thực hiện phép tính
a)\(\frac{45}{19}-\left(\frac{1}{2}+\left(\frac{1}{3}+\left(\frac{1}{4}\right)^{-1}\right)^{-1}\right)^{-1}.\)
b) \(\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^{10}.6^{19}-7.2^{29}.27^6}.\)
Bài 2. tìm x, biết:
a) 2(x-1) - 3(2x+2) - 4(2x+3) =16
b) \(3\frac{1}{2}:\left|2x-1\right|=\frac{21}{22}\)
c) |x2+|x-1|| = x2+2
Bài 3. Chứng minh rằng số có dạng abcabc luôn chia hết cho 11
Bài 4.tính:
a) A = \(\left(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}\right):\frac{2012}{2013}\)
b) B =\(4.\left(-\frac{1}{2}\right)^2-2.\left(-\frac{1}{2}\right)^2+3.\left(-\frac{1}{2}\right)+1\)
c) C =\(\frac{1}{2}:\left(-1\frac{1}{2}\right):1\frac{1}{3}:\left(-1\frac{1}{4}\right):1\frac{1}{5}:\left(-1\frac{1}{6}\right):...:\left(-1\frac{1}{100}\right)\)
d) D =\(\frac{4^6.9^5+6^9.120}{-8^4.3^{12}+6^{11}}\)
tìm x
a , \(2.3^{x+2}+4.3^{x+1}=3^6.10\)
b, \(\left(\frac{1}{3}+\frac{1}{6}\right).2^{x+4}-2^x=2^{13}-2^{16}\)
a, Ta có \(2.3^{x+2}+4.3^{x+1}=3^6.10\)
\(\Rightarrow2.3.3^{x+1}+4.3^{x+1}=3^6.10\)
\(\Rightarrow3^{x+1}.\left(6+4\right)=3^6.10\)
\(\Rightarrow3^{x+1}.10=3^6.10\)
\(\Rightarrow3^{x+1}=3^6\)
\(\Rightarrow x+1=6\)
\(\Rightarrow x=5\)
b,\(\left(\frac{1}{3}+\frac{1}{6}\right).2^{x+4}-2^x=2^{13}-2^{16}\)
\(\Rightarrow\frac{1}{2}.2^{x+4}-2^x=2^{13}.\left(1-2^3\right)\)
\(\Rightarrow2^{x+3}-2^x=2^{13}.\left(1-2^3\right)\)
\(\Rightarrow2^x.\left(2^3-1\right)=2^{13}.\left(1-2^3\right)\)
\(\Rightarrow2^x.\left(2^3-1\right)=-2^{13}.\left(2^3-1\right)\)
\(\Rightarrow2^x=2^{-13}\)
\(\Rightarrow x=-13\)
A ) 2 . 3x+2 + 4 . 33+1 = 36 . 10
2 . 3x . 9 + 4 . 3x . 3 = 729 .10
18 . 3x + 12 . 3x = 243 . 3 . 10
30 . 3x = 243 . 30
3x = 243
x = 5
tim x :
\(a.\left|x+5\right|=\frac{1}{7}-\left|\frac{4}{3}-\frac{1}{6}\right|\)|
\(b.\left|x+\frac{2}{3}\right|=\frac{1}{2}-\left[\frac{1}{4}-\frac{2}{3}\right]\)đau [ ] là dấu ngoặc đơn nha!
\(c.\left|x-\frac{1}{5}\right|=\left|\frac{-5}{2}\right|-\left[\frac{1}{4}-\frac{2}{3}\right]\)đau [ ] là dấu ngoặc đơn nha!
\(d.\left|x-\frac{5}{2}\right|=\frac{4}{3}-\left[\frac{2}{3}-\frac{1}{6}\right]\)đau [ ] là dấu ngoặc đơn nha!
ai nhanh nhất mk tk !
cau a dau nhi cuoi cung k phai j dau nha ! mk an lom !
\(a,\)\(\left|x+5\right|=\frac{1}{7}-\left|\frac{4}{3}-\frac{1}{6}\right|\)
\(\Leftrightarrow\left|x+5\right|=\frac{1}{7}-\frac{7}{6}\)
\(\Leftrightarrow\left|x+5\right|=\frac{-43}{42}\)
ta có |x+5| \(\ge\)0 \(\forall x\)
Mà \(-\frac{43}{42}< 0\)nên ko có giá trị x thoả mãn
b,
\(\left|x+\frac{2}{3}\right|=\frac{1}{2}-\left(\frac{1}{4}+\frac{2}{3}\right)\)
\(\Leftrightarrow\left|x+\frac{2}{3}\right|=\frac{11}{12}\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{2}{3}=\frac{11}{12}\forall x\ge-\frac{2}{3}\\-x-\frac{2}{3}=\frac{11}{12}\forall< -\frac{2}{3}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=-\frac{19}{12}\end{cases}}\)(thoả mãn đk)