(X-5) mũ 0 = (x-5) mũ 8
(X+5).(x-4)=0
(X-1).(x-3)=0
|5-x|=7
1.(x -5) mũ 2 - 25 =0
2. (x -2) mũ 3 =27
3. 3(x -7) + 2x(x+2) = 2x mũ 2
4. (x mũ 2 - 4) (x +8) =0
5. x mũ 2 + 3x = 0
6. 3x mũ 3 - 3x = 0
7. (x +1) mũ 2 = ( 2x +3) mũ 2
1.(x -5)^2 - 25 =0
=> (x - 5)^2 = 25
=> x - 5 = 5 hoặc x - 5 = -5
=> x = 10 hoặc x = 0
vậy_
2. (x -2)^3 =27
=> x - 2 = 3
=> x = 5
vậy_
3. 3(x -7) + 2x(x+2) = 2x^2
=> 3x - 21 + 2x^2 + 4x = 2x^2
=> 7x - 21 = 0
=> 7x = 21
=> x = 3
vậy_
4. (x^2 - 4) (x +8) =0
=> x^2 - 4 = 0 hoặc x + 8 = 0
=> x^2 = 4 hoặc x = -8
=> x = 2 hoặc x = -2 hoặc x = -8
vậy_
5. x^ 2 + 3x = 0
=> x(x + 3) = 0
=> x = 0 hoặc x + 3 = 0
=> x = 0 hoặc x = -3
vậy_
6. 3x^3 - 3x = 0
=> 3x(x^2 - 1) = 0
=> 3x(x - 1)(x + 1) = 0
=> x = 0 hoặc x = 1 hoặc x = -1
vậy_
7. (x +1)^2 = ( 2x +3)^2
=> (x + 1 + 2x + 3)(x + 1 - 2x - 3) = 0
=> (3x + 3)(-x - 2) = 0
=> x = -1 hoặc x = -2
vậy_
Bài làm
1) ( x - 5 )2 - 25 = 0
<=> ( x - 5 - 5 )( x - 5 + 5 ) = 0
<=> x( x - 10 ) =
<=> \(\orbr{\begin{cases}x=0\\x-10=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=10\end{cases}}}\)
Vậy S = { 0; 10 }
2) \(\left(x-2\right)^3=27\)
\(\Leftrightarrow\left(x-2\right)^3=3^3\)
\(\Leftrightarrow x-2=3\)
\(\Leftrightarrow x=5\)
Vậy x = 5 là nghiệm phương trình.
3) \(3\left(x-7\right)+2x\left(x+2\right)=2x^2\)
\(\Leftrightarrow3x+2x^2+4x-2x^2=21\)
\(\Leftrightarrow7x=21\)
\(\Leftrightarrow x=\frac{21}{7}=3\)
Vậy x = 3 là nghiệm phương trình
4) \(\left(x^2-4\right)\left(x+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-4=0\\x+8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2=\pm2\\x=-8\end{cases}}}\)
Vậy S = { 2; -2; -8 }
5) \(x^2+3x=0\)
\(\Leftrightarrow x\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-3\end{cases}}}\)
Vậy S = { 0; -3 }
6) \(3x^3-3x=0\)
\(\Leftrightarrow3x\left(x^2-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x=0\\x^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}}\)
Vậy S = { +1; 0 }
7) \(\left(x+1\right)^2=\left(2x+3\right)^2\)
\(\Leftrightarrow\left(x+1\right)^2-\left(2x+3\right)^2=0\)
\(\Leftrightarrow\left(x+1-2x-3\right)\left(x+1+2x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}-x-2=0\\3x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{4}{3}\end{cases}}}\)
Vậy S = { -2; -4/3 }
# Học tốt #
tìm x ,y bt (/ là giá trị tuyệt đối nhé)
a,/x-3/+/x+5/-8=0
b,/2x+1/+*2x-5/-4=0
c,/x-3/+/3x+4/+/2x-1/=8
d,/x-3y/ mũ 11 +(y+4) mũ 12=0
e,(x+y) mũ 2016 + 2017/y-1/ mũ 3 = 0
d,/x-y-5/+2015(y-3) mũ 2016=0
f,(x-1) mũ 2 + (y+3) mũ 4 = 0
g, 2(x-5) mũ 6 + 5[/2y-7/ mũ 5]=0
ch,/x=3y-1/+(3y-2) mũ 2016 =0
Nếu dc mọi người có thể chỉ rõ cho em cách giả dc ko ạ,lần sau có j em còn bt làm.Em cảm ơn ạ
1. 6 X mũ 3 -8 =40
2. 4 X mũ 5 +15=47
3. 2 X mũ 3-4=12
4. 5 X mũ 3-5=0
5. (X -5) mũ 2016 = (X-5) mũ 2018
6. (3X -2) mũ 20= (3X-1) mũ 20
7. (3X -1) mũ 10 = (3X-1) mũ 20
8. (2X -1) mũ 50 = 2X-1
9. (X phần 3 -5) mũ 2000= ( X phần 3-5) mũ 2008
1. \(6x^3-8=40\\ 6x^3=48\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2
2. \(4x^5+15=47\\ 4x^5=32\\ x^5=8\\ \Rightarrow x\in\varnothing\left(\text{vì }x\in N\right)\)Vậy x ∈ ∅
3. \(2x^3-4=12\\ 2x^3=16\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2
4. \(5x^3-5=0\\ 5x^3=5\\ x^3=1\\ \Rightarrow x=1\)Vậy x = 1
5. \(\left(x-5\right)^{2016}=\left(x-5\right)^{2018}\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\x-5=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)Vậy \(x\in\left\{5;6\right\}\)
6. \(\left(3x-2\right)^{20}=\left(3x-1\right)^{20}\\ \Rightarrow3x-2=3x-1\\ 3x-3x=2-1\\ 0=1\left(\text{vô lí}\right)\)Vậy x ∈ ∅
7. \(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\\ \left(3x-1\right)^{10}=\left[\left(3x-1\right)^2\right]^{10}\\ \Rightarrow\left(3x-1\right)^2=3x-1\\ \left(3x-1\right)^2-\left(3x-1\right)=0\\ \left(3x-1\right)\left[\left(3x-1\right)-1\right]=0\\ \left(3x-1\right)\left(3x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x-1=0\\3x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=1\\3x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{3}\left(\text{loại vì }x\in N\right)\\x=\frac{2}{3}\left(\text{loại vì }x\in N\right)\end{matrix}\right.\)Vậy x ∈ ∅
8. \(\left(2x-1\right)^{50}=2x-1\\ \left(2x-1\right)^{50}-\left(2x-1\right)=0\\ \left(2x-1\right)\left[\left(2x-1\right)^{49}-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}2x-1=0\\\left(2x-1\right)^{49}=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=1\\2x-1=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}\left(\text{loại vì }x\in N\right)\\x=1\left(t/m\right)\end{matrix}\right.\)Vậy x = 1
9. \(\left(\frac{x}{3}-5\right)^{2000}=\left(\frac{x}{3}-5\right)^{2008}\\ \left(\frac{x}{3}-5\right)^{2008}-\left(\frac{x}{3}-5\right)^{2000}=0\\ \left(\frac{x}{3}-5\right)^{2000}\left[\left(\frac{x}{3}-5\right)^8-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}\left(\frac{x}{3}-5\right)^{2000}=0\\\left(\frac{x}{3}-5\right)^8=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}-5=0\\\frac{x}{3}-5=1\\\frac{x}{3}-5=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}=5\\\frac{x}{3}=6\\\frac{x}{3}=4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\cdot3=15\\x=6\cdot3=18\\x=4\cdot3=12\end{matrix}\right.\)Vậy \(x\in\left\{15;18;12\right\}\)
\(1.6x^3-8=40\\ \Leftrightarrow6x^3=48\\ \Leftrightarrow x^3=8\Leftrightarrow x^3=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{2;-2\right\}\)
\(2.4x^3+15=47\) (T nghĩ đề là mũ 3)
\(\Leftrightarrow4x^3=32\Leftrightarrow x^3=8=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{2;-2\right\}\)
Câu 3, 4 tương tự nhé.
\(5.\left(x-5\right)^{2016}=\left(x-5\right)^{2018}\\ \Leftrightarrow\left(x-5\right)^{2018}-\left(x-5\right)^{2016}=0\\ \Leftrightarrow\left(x-5\right)^{2016}\left[\left(x-5\right)^2-1\right]=0\\ \Leftrightarrow\left(x-5\right)^{2016}\left(x-5-1\right)\left(x-5+1\right)=0\\ \Leftrightarrow\left(x-5\right)^{2016}\left(x-6\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-5\right)^{2016}=0\\x-6=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\x=6\\x=4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=6\\x=4\end{matrix}\right.\)
Vậy \(x\in\left\{4;5;6\right\}\)
/3 x + 2/ = /x - 8/
x.( 3 x + 6 )=0
( 2 x - 4 ).( x mũ 3 - 1 ) =0
( 4 x - 8 ) . ( x mũ 2 + 1 )=0
( x mũ 3 - 8 ) . ( 5 x + 10 )=0
/3 x - 10/ = 5
dấu / giá trị tuyệt đối nha các bạn
\(\left|3x+2\right|=\left|x-8\right|\)
\(\Leftrightarrow\orbr{\begin{cases}3x+2=x-8\\3x+2=8-x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x-x=-8-2\\3x+x=8-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=-10\\4x=6\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-5\\x=\frac{3}{2}\end{cases}}\)
\(x\left(3x+6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\3x+6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\3x=-6\end{cases}}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)
\(\left(2x-4\right)\left(x^3-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-4=0\\x^3-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=4\\x^3=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=1\end{cases}}\)
\(\left(4x-8\right)\left(5x+10\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}4x-8=0\\5x+10=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}4x=8\\5x=-10\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)
\(\left|3x-10\right|=5\)
\(\Leftrightarrow\orbr{\begin{cases}3x-10=5\\3x-10=-5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x=15\\3x=5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=\frac{5}{3}\end{cases}}\)
tìm x
3, ( x - 2 ) mũ 2 - 5( 2 - x ) = 0
4, ( x mũ 3 - 8 ) + 2x mũ 2 - 4x = 0
5, x mũ 2 ( x - 3 ) + 18 - 6x = 0
3, \(\left(x-2\right)^2-5\left(2-x\right)=0\Leftrightarrow\left(2-x\right)^2-5\left(2-x\right)=0\)
\(\Leftrightarrow\left(2-x-5\right)\left(2-x\right)=0\Leftrightarrow\left(x+3\right)\left(2-x\right)=0\Leftrightarrow x=-3;x=2\)
4, \(x^3-8+2x^2-4x=0\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)+2x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)^2=0\Leftrightarrow x=\pm2\)
5, \(x^2\left(x-3\right)+18-6x=0\Leftrightarrow x^2\left(x-3\right)-6\left(x-3\right)=0\)
\(\Leftrightarrow\left(x^2-6\right)\left(x-3\right)=0\Leftrightarrow x=\pm\sqrt{6};x=3\)
tìm x
3, ( x - 2 ) mũ 2 - 5( 2 - x ) = 0
x=-3, x=2
4, ( x mũ 3 - 8 ) + 2x mũ 2 - 4x = 0
x= 2 , x= -2
5, x mũ 2 ( x - 3 ) + 18 - 6x = 0
x=-căn bậc hai(6), x=căn bậc hai(6), x=3
sao mọi người tl mà ko k vậy
Tìm x:
a) (x-20) mũ 2 -(x+1)(x+3)=-7
b) (3x+5)(4-3x)=0
c) x mũ 3 -9x=0
d)2/3x (x mũ 2 -4)=0
e) (2x+1)-x(2x+1)=0
f)(2x-1) mũ 2 -(2x+5) (2x-5) =18
g)x mũ 2 -25 =6x-9
b, x = -5/3 hoặc x = 4/3.
c, x = 0 hoặc x = 3, -3.
d, x = 0 hoặc x = 2, -2.
e, x = 1 hoặc x = \(\dfrac{-1}{2}\).
a: \(\Leftrightarrow x^2-40x+400-x^2-4x-3=-7\)
=>-44x+397=-7
=>-44x=-404
hay x=101
b: \(\Leftrightarrow\left[{}\begin{matrix}3x+5=0\\4-3x=0\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{5}{3};\dfrac{4}{3}\right\}\)
c: \(\Leftrightarrow x\left(x^2-9\right)=0\)
=>x(x-3)(x+3)=0
hay \(x\in\left\{0;3;-3\right\}\)
d: \(\Leftrightarrow x\left(x-2\right)\left(x+2\right)=0\)
hay \(x\in\left\{0;2;-2\right\}\)
e: =>(2x+1)(1-x)=0
=>x=-1/2 hoặc x=1
Tìm x:
a) (x-20) mũ 2 -(x+1)(x+3)=-7
b) (3x+5)(4-3x)=0
c) x mũ 3 -9x=0
d)2/3x (x mũ 2 -4)=0
e) (2x+1)-x(2x+1)=0
f)(2x-1) mũ 2 -(2x+5) (2x-5) =18
g)x mũ 2 -25 =6x-9
1)Thực hiện phép tính:
(-23)+13+(-17)+57
(-26)+(-6)+(-75)+(-50)
14+6+(-9)+(-14)
(-123)+/-13/+(-7)
/0/+/45/+(-/-455)*/+/-796/
-/-33/+(-12)+18+/45-40/-57
(-8537)+(1975+8537)
273+[-34+27+(-273)]
-452-(-67+75-452)
2)Tìm x
1)89-(73-x)=20
2)(x+7)-25=13
198-(x+4)=120
140/(x-8)=7
x-[42+(-28)]=-8
x+5=20-(12-7)
(x-51)=2*2 mũ 3+20
4(x-3)=7 mũ 2-1 mũ 10
2 mũ x+1*2 mũ 2009=2 mũ 2010
2x-49=5*3 mũ2
3 mũ 2(x+4)-5 mũ 2=5* 2 mũ 2
6x+x=5 mũ 11/5 mũ 9+3
7x-x=5 mũ 21/5 mũ 19+3*2 mũ 2-1
.7x-2x=6 mũ 17/6 mũ 15+44/11
0/x=0
2 mũ x/2 mũ 5=1
/x-2/=0
/x-5/=7-(-3)
/x-5/=/-7/
/x/-5=3
15/x/=5
3)Tìm x biết;
1)24 ;36;160;chia hết cho x va x lớn nhất
2)64;48;88 chia hết cho x va x lớn nhất
3)x thuộc ƯC(54;12) và x lớn nhất
4)x chia hết cho 4;7;8 và x nhỏ nhất
5)x chia hết cho 2;3;5 và x nhỏ nhất
6)x thuộc BC(9;8) và x nhỏ nhất
4)Tìm x biết:
1)x thuộc ƯC(36;24) và x <20
2)x thuộc ƯC(60;84;120)và >6
Tìm x ( xoắn 3 đại số 8 )
1. x mũ 6 - 2x mũ 3 + 1 = 0
2. x mũ 6 + 1/4x mũ 3 + 1/64 = 0
3. | 2 - x | mũ 2 + 6x - 3 = 0
4. x mũ 3 - 10x mũ 2 + 25x = 0
5. 1/4x mũ 3 - 3x mũ 2 + 9x = 0
6. x mũ 5 - 16x = 0
7. 4x mũ 2 + 4x - 3 = 0
8. 4x mũ 2 + 28x + 48 = 0
9. 9x mũ 2 - 12x + 3 = 0
Các bạn giúp miik nhé, mik sẽ tick cho các bạn !!!!!!!
1. \(x^6-2x^3+1=0\Leftrightarrow\left(x^3-1\right)^2=0\Leftrightarrow x=1\)
2. \(x^6+\dfrac{1}{4}x^3+\dfrac{1}{64}=0\Leftrightarrow\left(x^3\right)^2+2.x^3.\dfrac{1}{8}+\left(\dfrac{1}{8}\right)^2=0\Leftrightarrow\left(x+\dfrac{1}{8}\right)^2=0\Leftrightarrow x=-\dfrac{1}{2}\)4. \(x^3-10x^2+25x=0\Leftrightarrow x^3-5x^2-5x^2+25x=0\)
\(\Leftrightarrow x^2\left(x-5\right)-5x\left(x-5\right)=0\)
\(\Leftrightarrow x\left(x-5\right)^2=0\Leftrightarrow x=5\)
5. \(\dfrac{1}{4}x^3-3x^2+9x=0\)
\(\Leftrightarrow x\left(\dfrac{1}{4}x^2-3x+9\right)=0\)
\(\Leftrightarrow x\left[\left(\dfrac{1}{2}x\right)^2-2.\dfrac{1}{2}x.3+3^2\right]=0\)
\(\Leftrightarrow x\left(\dfrac{1}{2}x-3\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
6. \(x^5-16x=0\Leftrightarrow x\left(x^4-16\right)=0\Leftrightarrow x\left(x^2-4\right)\left(x^2+4\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)\left(x^2+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\\x^2=-4\left(l\right)\end{matrix}\right.\)
7. \(4x^2+4x-3=0\Leftrightarrow4x^2-2x^2-6x-3=0\)
\(\Leftrightarrow2x\left(2x-1\right)-3\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
8. \(4x^2+28x+48=0\Leftrightarrow4x^2+12x+14x+48=0\)
\(\Leftrightarrow4x\left(x+3\right)+12\left(x+4\right)=0\)
\(\Leftrightarrow\left(4x+12\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-4\end{matrix}\right.\)
9. \(9x^2-12x+3=0\Leftrightarrow9x^2-9x-3x+3=0\Leftrightarrow9x\left(x-1\right)-3\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(9x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)
|2 - x|2 + 6x - 3 = 0
<=> (x - 2)2 + 6x - 3 = 0
<=> x2 - 4x + 4 + 6x - 3 = 0
<=> x2 + 2x + 1 = 0
<=> (x + 1)2 = 0
<=> x + 1 = 0
<=> x = -1
Bắt phải thể hiện -_-