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\(5^{x+4}-3.5^{x+3}=2.5^{11}\)
\(5^{x+3}\left(5-3\right)=2.5^{11}\)
\(5^{x+3}.2=2.5^{11}\)
\(5^{x+3}=5^{11}\)
\(x+3=11\)
\(x=8\)
\(4^{x+3}-3.4^{x+1}=13.4^{11}\)
\(4^{x+1}\left(4^2-3\right)=13.4^{11}\)
\(4^{x+1}.13=13.4^{11}\)
\(4^{x+1}=4^{11}\)
\(x+1=11\)
\(x=10\)
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
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)\(\left(\frac{4}{9}\right)^x=\left(\frac{8}{27}\right)^6\)
\(\left[\left(\frac{2}{3}\right)^2\right]^x=\left[\left(\frac{2}{3}\right)^3\right]^6\)
\(\left(\frac{2}{3}\right)^{2x}=\left(\frac{2}{3}\right)^{18}\)
\(\Rightarrow2x=18\)
\(\Rightarrow x=9\)
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
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)
Vế trái: 4/(x+2).(x+6)+7/(x+6).(x+13)
<=>1/x+2 -1/x+6 +1/x+6 -1/x+13
<=>1/x+2-1/x+13
=> 1/x+2-1/x+13=2x+1/(x+2).(x+16) -3/(x+13).(x+16)
<=>1/x+2 - 1/x+13 + 1/x+13 - 1/x+16=2x+1/(x+2).(x+16)
<=>1/x+2 - 1/x+16=2x+1/(x+2).(x+16)
<=> 14/(x+2).(x+16)= 2x+1/(x+2).(x+16)
<=> 2x+1=14
<=> 2x=14-1
<=> 2x=13
<=> x=13:2
<=> x=13/2
Vậy x=13/2
Chúc bạn học tốt
a, \(\left(\frac{1}{2}-\frac{1}{3}\right)\cdot6^x+6^{x+2}=6^{10}+6^7\)
\(\Leftrightarrow\frac{1}{6}\cdot6^x+6^x\cdot6^2=6^{10}+6^7\)
\(\Leftrightarrow6^{x-1}\left(1+6^3\right)=6^7\left(6^3+1\right)\)
\(\Leftrightarrow6^{x-1}=6^7\Leftrightarrow x-1=7\)
\(\Leftrightarrow x=8\)
b, \(\left(\frac{1}{2}-\frac{1}{6}\right)\cdot3^{x+4}-4\cdot3^x=3^{16}-4\cdot3^{13}\)
\(\Leftrightarrow\frac{1}{3}\cdot3^{x+4}-4\cdot3^x=3^{13}\left(3^3-4\right)\)
\(\Leftrightarrow3^x\cdot3^3-4\cdot3^x=3^{13}\left(3^3-4\right)\)
\(\Leftrightarrow3^x\left(3^3-4\right)=3^{13}\left(3^3-4\right)\)
\(\Leftrightarrow3^x=3^{13}\Leftrightarrow x=13\)
a. x=8
b. x=13
còn cách tính thì mình quên rồi vì minh học cái này lâu lắm rồi ko nhớ đc.