Tìm số tự nhiên x biết
32 mũ x<128 mũ 4
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2x . 4 = 128
2x = 128 : 4
2x = 32
2x = 2 . 2 . 2 . 2 . 2
2x = 25
x = 5
(2x + 1)3 = 125
(2x + 1)3 = 5 . 5 . 5
(2x + 1)3 = 53
2x + 1 = 5
2x = 5 - 1
2x = 4
x = 4 : 2
x = 2
x15 = x
x = 1
(x - 5)4 = (x - 5)6
x = 6
a) 5x+x+1=\(\dfrac{125}{25}\)
\(\leftrightarrow\) 52x+1 =51
\(\leftrightarrow\) 2x+1=1
\(\leftrightarrow\)2x=0
\(\leftrightarrow\) x=0
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16x<1284
<=> (24)x < (27)4
<=> 2 4x< 228
=> 4x < 28 => x < 7
Đáp số: x < 7
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
\(128^4=\left(2^7\right)^4=2^{28}=\left(2^5\right)^{5,6}=32^{5,6}\)
theo đề ta có
\(32^x< 128^4=32^{5,6}\Rightarrow x\in\left\{1;2;3;4;5\right\}\)