Bài 2; Tìm x
a,25.5n+2=625
b,9x-5.9x=729
c,(3x-2)4=0
d,2x+6.2=128
viết các số sau về dạng lũy thừa của 1 số tự nhiên:125;27;64;1296;1024;2401;43;8;25.125
tìm n là số tự nhiên,biết
2n=16 3n=81 2n-1=64 3n+2=27.81 25.5n-1=625
2n.8=128 3.5n=375 (3n)2=729 81≤3n≤729 (2n-1)4=81
(n-2)3=125 (n-1)5=32.243
bài khó quá ,giúp mình với
1. (Mình đưa nó về thừa số nguyên tố nha, cái nào ko đc thì thôi)
125 = 53; 27 = 33; 64 = 26; 1296 = 64; 1024 = 210; 2401 = 74; 43 = 64; 8 = 23; 25.125 = 3125 = 55.
2.
2n = 16 =) n = 4. 3n = 81 =) n = 4. 2n-1 = 64 =) n = 7. 3n+2 = 27.81 =) n = 5. 25.5n-1 = 625 =) n = 3.
2n.8 = 128 =) n = 4. 3.5n = 375 =) n = 3. (3n)2 = 729 =) n = 3. 81 ≤ 3n ≤ 729 =) n = 4; 5; 6.
\(125=5^3;27=3^3;1296=36^2=6^4=2^4.3^4;1024=32^2=2^{10};2401=49^2=7^4;4^3=2^6;8=2^3;25.125=5^2.5^3=5^5\)
Tìm x ∈ N , biết.
a) 3 x + 1 : 3 4 = 81
b) 3 x + 3 . 3 x + 1 = 729
c) 2 x + 3 . 2 x = 128
d) 23 + 3 x = 5 6 : 5 3
e) 2 x + 2 x + 4 = 272
Tìm x ∈ N, biết.
a, 3 x + 1 : 3 4 = 81
b, 3 x + 3 . 3 x + 1 = 729
c, 2 x + 3 . 2 x = 128
d, 23 + 3 x = 5 6 : 5 3
e, 2 x + 2 x + 4 = 272
a, 3 x + 1 : 3 4 = 81
3 x - 3 = 3 4
x – 3 = 4
x = 7
Vậy x = 7
b, 3 x + 3 . 3 x + 1 = 729
3 2 x + 4 = 3 6
2x + 4 = 6
x = 1
Vậy x = 1
c, 2 x + 3 . 2 x = 128
2 2 x + 3 = 2 7
2x + 3 = 7
x = 2
Vậy x = 2
d, 23 + 3 x = 5 6 : 5 3
23 + 3 x = 5 3
23 + 3x = 125
3x = 102
x = 34
Vậy x = 34
e, 2 x + 2 x + 4 = 272
2 x + 2 x . 2 4 = 272
2 x ( 1 + 2 4 ) = 272
2 x . 17 = 272
2 x = 16
2 x = 2 4
x = 4
Vậy x = 4
bài 2; tìm x
a, 5x ( x - 1 ) + ( x + 17 ) = 0
b, 3x ( x - 3 ) mũ 2 - 3x ( x + 3 ) mũ 2 = 0
c, 7 - 9x + 2x mũ 2 = 0
d, 7 - 9x + 2x mũ 2 = 0
a, \(5x\left(x-1\right)+\left(x+17\right)=0\)
\(\Leftrightarrow5x^2-5x+x+17=0\Leftrightarrow5x^2-4x+17=0\)
\(\Leftrightarrow5\left(x^2-\frac{4}{5}x\right)+17=0\Leftrightarrow5\left(x^2-2.\frac{2}{5}x+\frac{4}{25}-\frac{4}{25}\right)+17=0\)
\(\Leftrightarrow5\left(x-\frac{2}{5}\right)^2-\frac{4}{5}+17=0\Leftrightarrow5\left(x-\frac{2}{5}\right)^2+81\ge81>0\)
Vậy pt vô nghiệm
b, \(3x\left(x-3\right)^2-3x\left(x+3\right)^2=0\)
\(\Leftrightarrow3x\left[\left(x-3\right)^2-\left(x+3\right)^2\right]=0\)
\(\Leftrightarrow3x\left(x-3-x-3\right)\left(x-3+x+3\right)=0\Leftrightarrow x.2x=0\Leftrightarrow x=0\)
c, \(2x^2-9x+7=0\Leftrightarrow2x^2-7x-2x+7=0\)
\(\Leftrightarrow x\left(2x-7\right)-\left(2x-7\right)=0\Leftrightarrow\left(x-1\right)\left(2x-7\right)=0\Leftrightarrow x=1;x=\frac{7}{2}\)
Trả lời:
a, \(5x\left(x-1\right)+\left(x+17\right)=0\)
\(\Leftrightarrow5x^2-5x+x+17=0\)
\(\Leftrightarrow5x^2-4x+17=0\)
\(\Leftrightarrow5\left(x^2-\frac{4}{5}x+\frac{17}{5}\right)=0\)
\(\Leftrightarrow x^2-\frac{4}{5}x+\frac{17}{5}=0\)
\(\Leftrightarrow x^2-2.x.\frac{2}{5}+\frac{4}{25}+\frac{81}{25}=0\)
\(\Leftrightarrow\left(x-\frac{2}{5}\right)^2+\frac{81}{25}=0\)
Vì \(\left(x-\frac{2}{5}\right)^2+\frac{81}{25}\ge\frac{81}{25}>0\forall x\)
nên pt vô nghiệm
b, \(3x\left(x-3\right)^2-3x\left(x+3\right)^2=0\)
\(\Leftrightarrow3x\left[\left(x-3\right)^2-\left(x+3\right)^2\right]=0\)
\(\Leftrightarrow3x\left(x-3-x-3\right)\left(x-3+x+3\right)=0\)
\(\Leftrightarrow3x.\left(-9\right).2x=0\)
\(\Leftrightarrow-54x^2=0\)
\(\Leftrightarrow x^2=0\)
\(\Leftrightarrow x=0\)
Vậy x = 0 là nghiệm của pt.
c, \(7-9x+2x^2=0\)
\(\Leftrightarrow2x^2-7x-2x+7=0\)
\(\Leftrightarrow x\left(2x-7\right)-\left(2x-7\right)=0\)
\(\Leftrightarrow\left(2x-7\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-7=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=1\end{cases}}}\)
Vậy x = 7/2; x = 1 là nghiệm của pt.
d, trùng ý c
10 Phân tích các đa thức sau thành nhân tử:
a) 5xy(x-y)-2x+2y ; b) 6x-2y-x(y-3x)
c) x^2+4x-xy-4y ; d) 3xy+2z-6y-xz
11 Tìm x, biết: a) 4-9x^2=0 ; b) x^2+x+1/4=0 ; c) 2x(x-3)+(x-3)=0
d) 3x(x-4)-x+4=0 ; e) x^3-1/9x=0 ; f) (3x-y)^2-(x-y)^2=0
a) 5xy ( x - y ) - 2x + 2y
= 5xy ( x - y ) - 2 ( x - y )
= ( x - y ) ( 5xy - 2 )
b) 6x-2y-x(y-3x)
= 2 ( y - 3x ) - x ( y - 3x )
= ( y - 3x ( ( 2 - x )
c) x2 + 4x - xy-4y
= x ( x + 4 ) - y ( x + 4 )
( x + 4 ) ( x - y )
d) 3xy + 2z - 6y - xz
= ( 3xy - 6y ) + ( 2z - xz )
= 3y ( x - 2 ) + z ( x - 2 )
= ( x - 2 ) ( 3y + z )
a,5xy(x-y)-2x+2y=5xy(x-y)-2(x-y)=(x-y)(5xy-2)
b,6x-2y-x(y-3x)=-2(y-3x)-x(y-3x)=(y-3x)(-2-x)
c,x^2+4x-xy-4y=x(x+4)-y(x+4)=(x+4)(x-y)
d,3xy+2z-6y-xz=(3xy-6y)+(2z-xz)=3y(x-2)+z(2-x)=3y(x-2)-z(x-2)=(x-2)(3y-z)
11)
a,4-9x^2=0
(2-3x)(2+3x)=0
2-3x=0=>x=2/3 hoặc 2+3x=0=>x=-2/3
b,x^2 +x+1/4=0
(x+1/2)^2 =0
x+1/2=0
x=-1/2
c,2x(x-3)+(x-3)=0
(x-3)(2x+1)=0
x-3=0=>x=3 hoặc 2x+1=0=>x=-1/2
d,3x(x-4)-x+4=0
3x(x-4)-(x-4)=0
(x-4)(3x-1)=0
x-4=0=>x=4 hoặc 3x-1=0=>x=1/3
e,x^3-1/9x=0
x(x^2-1/9)=0
x(x+1/3)(x-1/3)=0
x=0 hoặc x+1/3=0=>x=-1/3 hoặc x-1/3=0=>x=1/3
f,(3x-y)^2-(x-y)^2 =0
(3x-y-x+y)(3x-y+x-y)=0
2x(4x-2y)=0
4x(2x-y)=0
x=0hoặc 2x-y=0=>x=y/2
bài 1:chứng minh rằng với mọi x ta có:
a)-x^2+4x-5<0
b)x^4+3x^2+3>0
c)(x^2+2x+3)(x^2+2x+4)+3>0
bài 2:tìm x:
a)9x^2-6x-3=0
b)x^3+9x^2+27x+19=0
c)x(x-5)(x+5)-(x+2)(x^2-2x+4)=3
Bài 1:
a)-x^2+4x-5
=-(x2-4x+5)<0 với mọi x
=>-x^2+4x-5<0 với mọi x
b)x^4+3x^2+3
\(=\left(x^2+\frac{3}{2}\right)^2+\frac{3}{4}>0\)với mọi x
=>x^4+3x^2+3>0 với mọi x
c) bn xét từng th ra
Bài 2:
a)9x^2-6x-3=0
=>3(3x2-2x-1)=0
=>3x2-2x-1=0
=>3x2+x-3x-1=0
=>x(3x+1)-(3x+1)=0
=>(x-1)(3x+1)=0
b)x^3+9x^2+27x+19=0
=>(x+1)(x2+8x+19) (dùng pp nhẩm nghiệm rồi mò ra)
Với x+1=0 =>x=-1Với x2+8x+19 =>vô nghiệmc)x(x-5)(x+5)-(x+2)(x^2-2x+4)=3
=>x3-25x-x3-8=3
=>-25x-8=3
=>-25x=1
=>x=-11/25
mk sửa 1 tí ở dấu => thứ 2 từ dưới lên là
=>-25x=11
Bài 7. Tìm số tự nhiên x biết:
a) 3x = 243
b) 56 : 5x = 625
c) 2x .24 = 128
d) 2.3x+2 = 18
mink sẽ tíck cho những ai chả lời cho mink!
a)x=243:3 b) 5x=56:625 c)2x=128:24 d)3x+2=18:2
x=81 5x=28/263 2x=16/3 3x+2=9
x=28/263:5 x=16/3:2 3x=9-2
x=28/1315 x=8/3 3x=7=>x=7/3
a) 3x = 243
x = 243 : 3
x = 81
b ) 56 : 5x = 625
5x = 56 : 625
5x = 56/625
x = 56/625 : 5
x = 56/3125
Tìm x
a.(x+2).(x+3)-(x-2).(x+5) = 0
b. (2x+3).(x-4)+(x-5)(x+2) = (3x-5)(x-4)
c. (3x+2)(2x+9)-(x+2)(6x+1) = x+1-(x-6)
d. 3( 2x-1).(3x-1)-(2x-3).(9x-1)=0
(x+2)(x+3)-(x-2)(x+5)=0
=> x2+5x+6-x2-3x+10=0
=>2x+16=0
=>2x=-16
=>x=-8
Tìm x biết:
a,(2x+3/5)^2-9/25=0
b,(3x-1).(-1/2x+5)=0
c, (7/5)^x+1-(1/5)^x=-4/625
d,(2/3)^x+2+(2/3)^x+1=20/27
a) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(=>2x+\frac{3}{5}=\frac{3}{5}\)
\(2x=\frac{3}{5}-\frac{3}{5}\)
\(2x=0\)
\(x=0:2\)
\(x=0\)
b) \(\left(3x-1\right).\left(-\frac{1}{2x}+5\right)=0\)
=> \(\left(3x-1\right)=0\)hoặc \(\left(-\frac{1}{2x}+5\right)=0\)hoặc \(\left(3x-1\right)\)và\(\left(-\frac{1}{2x}+5\right)\)cùng bằng 0.
\(\orbr{\begin{cases}3x-1=0\\-\frac{1}{2x}+5=0\end{cases}}=>\orbr{\begin{cases}3x=1\\-\frac{1}{2x}=-5\end{cases}}=>\orbr{\begin{cases}x\in\varnothing\\2x=\frac{1}{5}\end{cases}}=>x=\frac{1}{5}:2=>x=\frac{1}{10}\)
a) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\left(2x+\frac{3}{5}\right)^2=0+\frac{9}{25}\)
\(\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\orbr{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)
\(\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)
b) \(\left(3x-1\right)\left(-\frac{1}{2}x+5\right)=0\)
\(\left(3x-1\right)\left(-\frac{x}{2}+5\right)=0\)
\(\left(3x-1\right)\left(5-\frac{x}{2}\right)=0\)
\(\orbr{\begin{cases}3x-1=0\\5-\frac{x}{2}=0\end{cases}}\)
\(\orbr{\begin{cases}x=\frac{1}{3}\\x=10\end{cases}}\)