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2x + 2 + 2x - 1 + 2x - 2 = 152
19 . 2x - 2 = 152
19 . 2x - 2/19 = 152/19
2x - 2 = 8
2x - 2 = 23
x - 2 = 3
x = 3 + 2
x = 5
Bài 1 :
a, \(\left(x^2-29\right)^3=343\)
=> \(\left(x^2-29\right)^3=7^3\)
=> \(x^2-29=7\)
=> \(x^2=7+29=36\)
=> \(\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)
Do x là số tự nhiên => x = 6
b, \(2^{x+2}+2^{x-1}+2^{x-2}=152\)
=> \(2^x.2^2+2^x:2^1+2^x:2^2=152\)
=> \(2^x.2^2+2^x.\frac{1}{2}+2^x.\frac{1}{4}=152\)
=> \(2^x.\left(2^2+\frac{1}{2}+\frac{1}{4}\right)=152\)
=> \(2^x.\frac{19}{4}=152\)
=> \(2^x=32\)
=> \(2^x=2^5\)
=> x = 5
Bài 2 :
a, \(\left(2^9.76+2^{10}.35\right).3=2^{10}.38+2^{10}.35=2^{10}\left(38+35\right).3=2^{10}.73.3=1024.3.73=224256\)
b, \(\frac{\left(2^9.76+2^{10}.35\right).3}{2^8.438}=\frac{2^{10}.73.3}{2^9.219}=\frac{2^{10}.219}{2^9.219}=2\)
Bài 1 . Tìm x
a) 723 - ( 7x - 152 ) = 714
7x - 152 = 723 - 714
7x - 152 = 9
7x = 9 + 152
7x = 161
x = 161 : 7
x = 23
Vậy x = 23
b) ( 2x - 130 ) : 4 + 213 = 52 + 193
( 2x - 130 ) : 4 + 213 = 218
( 2x - 130 ) : 4 = 218 - 213
( 2x - 130 ) : 4 = 5
2x - 130 = 5 . 4
2x - 130 = 20
2x = 20 + 130
2x = 150
x = 150 : 2
x = 75
Vậy x = 75
c) ( x - 6 )2 = 9
( x - 6 )2 = 32
x - 6 = 3 <=> x = 3 + 6 <=> x = 9
x - 6 = -3 <=> x = -3 + 6 <=> x = 3
a) -152 - (3x + 1) = (-2).(-3)3
-152 - 3x - 1 = (-2).(-27)
-3x - 153 = 54
-3x = 54 + 153
-3x = 207
x = -69
b) x - 43 = (35 - x) - 48
x - 43 = 35 - x - 48
x - 43= -x - 13
x = -x - 13 + 43
x = 30 + x
x + x = 30
2x = 30
x = 15
Giải:
a) \(A=17-\left(x+4\right)^2\le17;\forall x\)
\(\Leftrightarrow A_{Max}=17\)
\("="\Leftrightarrow x+4=0\Leftrightarrow x=-4\)
Vậy ...
b) \(B=-x^2+8x+152\)
\(\Leftrightarrow B=-x^2+8x-16+168\)
\(\Leftrightarrow B=-\left(x^2-8x+16\right)+168\)
\(\Leftrightarrow B=-\left(x-4\right)^2+168\)
\(\Leftrightarrow B=168-\left(x-4\right)^2\le168;\forall x\)
\(\Leftrightarrow B_{Max}=168\)
\("="\Leftrightarrow x-4=0\Leftrightarrow x=4\)
Vậy ...
Bài 1:
1, ( 32 )3 x 272 x 93
=36*(33)2*(32)3
=36*36*36
=36+6+6
=318
2, 43 x 82 x (25 ) 2 x 32
=(22)3*(23)2*25*(52)2
=26*26*25*54
=26+6+5*54
=217*54 nếu đề ko sai thì đây là tối giản r
3, 62 x 36 x ( 62) 5
=62*62*610
=62+2+10
=614
Bài 2:phân tích tương tự nhưng khó hơn 1 tí
d)
\(152-3.\left\{\left[32-2\left(36\div x^2+7\right)\right]+42\right\}=-4\)
\(\Rightarrow3.\left\{\left[32-2\left(36\div x^2+7\right)\right]+42\right\}=152-\left(-4\right)\)
\(\Rightarrow3.\left\{\left[32-2\left(36\div x^2+7\right)\right]+42\right\}=156\)
\(\Rightarrow\left[32-2\left(36\div x^2+7\right)\right]+42=156\div3\)
\(\Rightarrow\left[32-2\left(36\div x^2+7\right)\right]+42=52\)
\(\Rightarrow32-2\left(36\div x^2+7\right)=52-42\)
\(\Rightarrow32-2\left(36\div x^2+7\right)=10\)
\(\Rightarrow2\left(36\div x^2+7\right)=32-10\)
\(\Rightarrow2\left(36\div x^2+7\right)=22\)
\(\Rightarrow36\div x^2+7=22\div2\)
\(\Rightarrow36\div x^2+7=11\)
\(\Rightarrow36\div x^2=11-7\)
\(\Rightarrow36\div x^2=4\)
\(\Rightarrow x^2=36\div4\)
\(\Rightarrow x^2=9\)
\(\Rightarrow x^2=3^2\)
\(\Rightarrow x=3\)
Vậy \(x=3\)
Đặt \(2^x=t\left(t>0\right)\)
\(\Rightarrow4t+\frac{1}{2}t+\frac{1}{4}t=152\)
\(\Rightarrow t=32\Rightarrow2^x=32\Rightarrow x=5\)
2x+2 + 2x-1 + 2x-2 = 152
2x.4 + 2x.1/2 + 2x.1/4 = 152
2x.(4+1/2+1/4) = 152
2x.19/4 = 152
2x = 32 = 25
=> x = 5