TÌM GTLN CỦA BT SAU
C = 1/|x-2|+3
TÌM X
a, (1/2)^3x-1 = 1/32
b, 2*3^x-405= 3^x-1
c, (1/81)*27^2x=(-9)^4
d, (4x-1)^30=(4x-1)^20
: Tìm x, biết:
a) 3x( 4x- 1) - 2x(6x- 3 )=30 b) 2x(3-2x) + 2x(2x-1)=15
c) (5x-2)(4x-1) + (10x +3)(2x - 1)=1 d) (x+2) (x+2)- (x -3)(x+1) = 9
e) (4x+1)(6x-3) = 7 + (3x – 2)(8x + 9) g) (10x+2)(4x- 1)- (8x -3)(5x+2) =14
`@` `\text {Ans}`
`\downarrow`
`a)`
`3x(4x-1) - 2x(6x-3) = 30`
`=> 12x^2 - 3x - 12x^2 + 6x = 30`
`=> 3x = 30`
`=> x = 30 \div 3`
`=> x=10`
Vậy, `x=10`
`b)`
`2x(3-2x) + 2x(2x-1) = 15`
`=> 6x- 4x^2 + 4x^2 - 2x = 15`
`=> 4x = 15`
`=> x = 15/4`
Vậy, `x=15/4`
`c)`
`(5x-2)(4x-1) + (10x+3)(2x-1) = 1`
`=> 5x(4x-1) - 2(4x-1) + 10x(2x-1) + 3(2x-1)=1`
`=> 20x^2-5x - 8x + 2 + 20x^2 - 10x +6x - 3 =1`
`=> 40x^2 -17x - 1 = 1`
`d)`
`(x+2)(x+2)-(x-3)(x+1)=9`
`=> x^2 + 2x + 2x + 4 - x^2 - x + 3x + 3=9`
`=> 6x + 7 =9`
`=> 6x = 2`
`=> x=2/6 =1/3`
Vậy, `x=1/3`
`e)`
`(4x+1)(6x-3) = 7 + (3x-2)(8x+9)`
`=> 24x^2 - 12x + 6x - 3 = 7 + (3x-2)(8x+9)`
`=> 24x^2 - 12x + 6x - 3 = 7 + 24x^2 +11x - 18`
`=> 24x^2 - 6x - 3 = 24x^2 + 18x -11`
`=> 24x^2 - 6x - 3 - 24x^2 + 18x + 11 = 0`
`=> 12x +8 = 0`
`=> 12x = -8`
`=> x= -8/12 = -2/3`
Vậy, `x=-2/3`
`g)`
`(10x+2)(4x- 1)- (8x -3)(5x+2) =14`
`=> 40x^2 - 10x + 8x - 2 - 40x^2 - 16x + 15x + 6 = 14`
`=> -3x + 4 =14`
`=> -3x = 10`
`=> x= - 10/3`
Vậy, `x=-10/3`
Tìm bt lớn nhất hay nhỏ nhất nếu có của các bt sau:
a) x^2 + x + 2/3
b)9x^2 - 2x - 1/3
c)5x^2 - 2x + 1
d)-x^2 + 3x - 1
e)-4^2 - 6x + 3
f)-3x^2 + 4x - 1/2
g)x^2 + 2x - 1
h)x^2 - 6x + 9
i)4x^2 - 2x
a) Ta có : \(x^2+x+\frac{2}{3}\)
\(=x^2+2.x.\frac{1}{2}+\frac{1}{4}+\frac{5}{12}\)
\(=\left(x^2+2.x.\frac{1}{2}+\frac{1}{4}\right)+\frac{5}{12}\)
\(=\left(x+\frac{1}{2}\right)^2+\frac{5}{12}\)
Mà ; \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)
Nên : \(\left(x+\frac{1}{2}\right)^2+\frac{5}{12}\ge\frac{5}{12}\forall x\)
Vậy GTNN của biểu thức là : \(\frac{5}{12}\) khi \(x=-\frac{1}{2}\)
tìm x, biết:
a. 2 . 3x - 405 = 3^x - 1
b.(1/81)^x . 27^2x = (-9)^4
C/m gt bt sau ko phụ thuộc vào gtri của biến
a) (x+3)(x^2 - 3x+9)-(54+x^3)
b (2x+3)(4x^2 - 6x+9)-2(4x^3 -1)
c (x+3)^3 - (x+9)(x^2 + 27)
giúp mình với
a)9x2 – 49 = 0
b)(x – 1)(x + 2) – x – 2 = 0
c)(4x + 1)(x - 2) - (2x -3)(2x + 1) = 7
d)x(3x + 2) + (x + 1)2 – (2x – 5)(2x + 5) = 0
e)(x + 3)(x2 – 3x + 9) –x(x – 1)(x + 1) – 27 = 0
f)(4x-3)^2-3x(3-4x)=0
\(a,\Leftrightarrow\left(3x-7\right)\left(3x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{7}{3}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-1-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\\ c,\Leftrightarrow4x^2-7x-2-4x^2+4x+3=7\\ \Leftrightarrow-3x=6\Leftrightarrow x=-2\\ d,\Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25=0\\ \Leftrightarrow4x=-26\Leftrightarrow x=-\dfrac{13}{2}\\ e,\Leftrightarrow x^3+27-x^3+x-27=0\\ \Leftrightarrow x=0\\ f,\Leftrightarrow\left(4x-3\right)\left(4x-3+3x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)
a) 9x2-49=0
(3x)2-72=0
<=> (3x-7)(3x+7)=0
th1: 3x-7=0
<=>3x=7
<=>x=\(\dfrac{7}{3}\)
th2: 3x+7=0
<=>3x=-7
<=>x=\(-\dfrac{7}{3}\)
tìm x biết
a,\( |2-\frac{3}{2}x|-4=x+2\)
b,\(\left(4x-1\right)^{30}=\left(4x-1\right)^{20}\)
c,\(2.3^x-405=3^{x-1}\)
d.\(\left(\frac{1}{81}\right)^x.27^{2.x}=\left(-9\right)^4\)
e,\(\frac{x+1}{65}+\frac{x+3}{63}=\frac{x+5}{61}+\frac{x+7}{59}\)
g,\(\left(\frac{1}{2}\right)^{3x-1}=\frac{1}{32}\)
h,\(|3x-5|=|\frac{1}{2}x+3|\)
i,\(\left(\frac{3}{4}-x\right)^3=\frac{1}{64}\)
a) \(\left|2-\frac{3}{2}x\right|-4=x+2\)
=> \(\left|2-\frac{3}{2}x\right|=x+2+4\)
=> \(\left|2-\frac{3}{2}x\right|=x+6\)
ĐKXĐ : \(x+6\ge0\) => \(x\ge-6\)
Ta có: \(\left|2-\frac{3}{2}x\right|=x+6\)
=> \(\orbr{\begin{cases}2-\frac{3}{2}x=x+6\\2-\frac{3}{2}x=-x-6\end{cases}}\)
=> \(\orbr{\begin{cases}2-6=x+\frac{3}{2}x\\2+6=-x+\frac{3}{2}x\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{5}{2}x=-4\\\frac{1}{2}x=8\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{8}{5}\\x=16\end{cases}}\) (tm)
b) \(\left(4x-1\right)^{30}=\left(4x-1\right)^{20}\)
=> \(\left(4x-1\right)^{30}-\left(4x-1\right)^{20}=0\)
=> \(\left(4x-1\right)^{20}.\left[\left(4x-1\right)^{10}-1\right]=0\)
=> \(\orbr{\begin{cases}\left(4x-1\right)^{20}=0\\\left(4x-1\right)^{10}-1=0\end{cases}}\)
=> \(\orbr{\begin{cases}4x-1=0\\\left(4x-1\right)^{10}=1\end{cases}}\)
=> \(\orbr{\begin{cases}4x=1\\4x-1=\pm1\end{cases}}\)
=> x = 1/4
hoặc x = 0 hoặc x = 1/2
1) Chứng minh bt sau ko phụ thuộc vào biến
a) ( x-1)^ 3 - ( x+4) ( x^2- 4x+16) + 3x ( x-1)
b) (2x+3y) ( 4x^2- 6xy + 9y^2) - ( 2x - 3y ) ( 4x^2+ 6xy + 9y^2) - 27 ( 2y^3- 1 )
c) y( x^2- y^2) ( x^2+ y^2) - y( x^4- y^4)
d) ( x-1)^3- ( x-1) ( x^2+ x + 1 ) - 3 ( 1-x).x
1) tìm x :
a) (x-3)(x+3)-(x+5)(x-1)=6
b)(3x-2)^2-(2x+1)^2-(5x-1)(x+1)=20
c)(2x+1)(4x^2-2x+1)+(3-2x)(9+6x+4x^2)+12x=4
Bài 3.
1.A=(x-3).(x+3)+15-x^2
2. B=(x-1).(x^2+x+1)-x.(x^2+2)+2x
3. C=92x-1).(4x^2+2x+1 ) -x.(8x2 +1)+x
4.(2x-3).(4x^2+6x+9)-2x.(4x^2-1)=2x-27
5. (x-3).(x^2+3x+9)=x^3-27
ghi rõ cách làm nha
1.A =( x-3)( x+3) + 15 - x2
A=X2-3X+3X+15-X3
A=15-X
2.B=(X -1) (X2+X+1) - X (X2+2) + 2X
B=X3+ X2+ X - X2 - X - 1 - X3 - 2X + 2X
B= -1
3.C=(2X - 1 ) (4X2 + 2X + 1) - X ( 8 X 2 + 1 ) + X
C=8X3 - 4X2 +4X2 - 2X +2 X - 1 - 8X22 - X + X
C=8X3 - 1 - 8X22
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