I2xI=3x+4
?
Tìm x:
a,I2xI=3-x
b,Ix-1I=2x-1
c,I9-xI-9=3x
d,I3x-1I+2=x
e,I3x-5I+x=2
a) \(\left|2x\right|=3-x\)
\(\Rightarrow\orbr{\begin{cases}2x=3-x\\2x=x-3\end{cases}}\Rightarrow\orbr{\begin{cases}2x+x=3\\2x-x=-3\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x=3\\x=-3\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}\)
b) \(\left|x-1\right|=2x-1\)
\(\Rightarrow\orbr{\begin{cases}x-1=2x-1\\x-1=1-2x\end{cases}}\Rightarrow\orbr{\begin{cases}x-2x=-1+1\\x+2x=1+1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}-x=0\\3x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{2}{3}\end{cases}}\)
Mình làm mẫu câu a) nhé
Do |2x|>hoặc =0
=>3-x.hoặc =0
=>x<hoặc =3 (1)
Mà |2x| chẵn với mọi x
=>3-x là số chẵn
=>x lẻ (2)
Từ (1) và (2) ta có :
x thuộc {1;3}
+Nếu x=1=>|2x|=2
3-x=2 (t/mãn)
+Nếu x=3=>|2x|=6
3-x=0 (loại)
Vậy x =1
Giải pt
a. Ix+1I = x-2
b. Ix-1I = I2xI
c. Ix-3I + Ix-2I = 4
a) Ta có : Ix + 1I = x - 2
\(\Leftrightarrow\orbr{\begin{cases}x+1=x-2\\x+1=2-x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-x=-2-1\\x+x=2-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}0x=-3\\2x=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}0x=-3\left(v\text{ô}l\text{í}\right)\\x=\frac{1}{2}\end{cases}}\)
Giải pt
a. Ix+1I = x-2
b. Ix-1I = I2xI
c. Ix-3I + Ix-2I = 4
a) Ix + 1I = x - 2
<=> x + 1 = x - 2 hay x + 1 = 2 - x
<=> x - x = -2 - 1 I <=> x + x = 2 - 1
<=> 0x = -3 (vô lí) I <=> 2x = 1
I <=> x = 1/2
b) Ix - 1I = I2xI (*)
x | 0 | 1 | |||
x - 1 | - | - | - | 0 | + |
2x | - | 0 | + | + | + |
TH1: x < 0
(*) <=> 1 - x = -2x
<=> -x + 2x = -1
<=> x = -1
TH2: 0 <= x < 1
(*) <=> 1 - x = 2x
<=> -x - 2x = -1
<=> - 3x = -1
<=> x = 1/3
TH3: x >= 1
(*) <=> x - 1 = 2x
<=> x - 2x = 1
<=> -x = 1
<=> x = -1
c) Ix - 3I + Ix - 2I = 4 (**)
x | 2 | 3 | |||
x - 2 | - | 0 | + | + | + |
x - 3 | - | - | - | 0 | + |
TH1: x < 2
(**) <=> 3 - x + 2 - x = 4
<=> -2x = 4 - 3 - 2
<=> -2x = -1
<=> x = 1/2
TH2: 2 <= x < 3
(**) <=> 3 - x + x - 2 = 4
<=> 0x = 4 + 2 + 3
<=> 0x = 9 (vô lí)
TH3: x >= 3
(**) <=> x - 3 + x - 2 = 4
<=> 2x = 4 + 2 + 3
<=> 2x = 9
<=> x = 9/2
rút gọn b=(3x+4)^2+(5-3x)^2+2(3x+4)(5-3x)
\(\left(3x+4\right)^2+\left(5-3x\right)^2+2.\left(3x+4\right)\left(5-3x\right)\\ =\left[\left(3x+4\right)+\left(5-3x\right)\right]^2\\ =\left(3x+4+5-3x\right)^2\\ =9^2=81\)
Kết quả rút gọn biểu thức (3x+2).(3x-2) là A) 3x^2+4 B)3x^2-4 C)9x^2+4. D)9x^2-4
\(\left(3x+2\right)\left(3x-2\right)=9x^2-4\)
-> chọn D
x(1-3x)(4-3x)-(x-4)(3x+5)
x(1-3x)(4-3x)-(x-4)(3x+5)=(x-3x2)(4-3x)-(x-4)(3x+5)
=(4(x-3x2)-3x(x-3x2))-(3x(x-4)+5(x-4))
= (4x-12x2-3x2+9x3)-(3x2-12x+5x-20)
= (9x3-15x2+4x)-(3x2-7x-20)
= 9x3-15x2+4x-3x2-7x-20
= 9x3-18x2+x-20
p(x)=-2x+12x^2+3x^4-3x^2-3
g(x)=3x^4+x^2-4x^2+1,5x^2-3x^4+2x+1
thu gọn
\(P\left(x\right)=3x^4+9x^2-2x-3\)
\(Q\left(x\right)=\left(3x^4-3x^4\right)+\left(x^2-4x^2+1.5x^2\right)+2x+1=-1.5x^2+2x+1\)
B) (2x+3)2-(5x-4) (5x+4)=(x+5)2-(3x-1) (7x+2)-(x2-x+1)
C) (1-3x)2-(x-2) (9x+1)=(3x-4) (3x+4)-9(x+3)2
D) (3x+4) (3x-4) - (2x+5)2=(x-5)2+(2x+1)2-(x2-2x)+(x-1)2 cần gắp