a, \(\left|x-1,5\right|+\)\(\left|2,5-x\right|=0\)
b, \(\left|2x+3\right|-\)\(\left|x-4\right|=0\)
c, \(\left|x+1,2\right|+\)\(\left|x+1,3\right|=3x\)
Bài 1:
a) \(\left(x-1,3\right)^2=9\)
b) \(2^{4-x}=32\)
c) \(\left(x+1,5\right)^2+\left(y-2,5\right)^{10}=0\)
a) \(\left(x-1,3\right)^2=9\Leftrightarrow\left[{}\begin{matrix}x-1,3=3\\x-1,3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4,3\\x=-1,7\end{matrix}\right.\)
b) 24-x = 32
⇔ 24-x = 25
⇔ 4-x=5
⇔ x=-1
c) (x+1,5)2+(y-2,5)10=0
\(\Leftrightarrow\left\{{}\begin{matrix}x+1,5=0\\y-2,5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1,5\\y=2,5\end{matrix}\right.\)
\(a,\left(x-1,3\right)^2=9\\ \Leftrightarrow\left(x-1,3+9\right)\left(x-1,3-9\right)=0\\ \Leftrightarrow\left(x-7,7\right)\left(x-10,3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=7,7=\dfrac{77}{10}\\x=10,3=\dfrac{103}{10}\end{matrix}\right.\)
\(b,2^{4-x}=32=2^5\\ \Leftrightarrow4-x=5\\ \Leftrightarrow x=-1\)
\(c,\left(x+1,5\right)^2+\left(y-2,5\right)^{10}=0\\ \Leftrightarrow\left\{{}\begin{matrix}x+1,5=0\\y-2,5=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-1,5=-\dfrac{3}{2}\\y=2,5=\dfrac{5}{2}\end{matrix}\right.\)
a. (x - 1,3)2 = 9
<=> (x - 1,3)2 - 9 = 0
<=> (x - 1,3)2 - 32 = 0
<=> (x - 1,3 - 3)(x - 1,3 + 3) = 0
<=> (x - 4,3)(x + 1,7) = 0
<=> \(\left[{}\begin{matrix}x-4,3=0\\x+1,7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4,3\\x=-1,7\end{matrix}\right.\)
tim x
a,\(\left|2x-1,6\right|-2,3=1,4\)
b\(5,4-\left|3x-1,2\right|=5,5\)
c,\(\left|x+1,3\right|+\left|x+2,4\right|=4x\)
d, \(\left|x-1,2\right|+\left|2,5-x\right|=0\)
giup minh nhe minh dang can gap
a/ \(\left|2x-1,6\right|-2,3=1,4\)
\(\Leftrightarrow\left|2x-1,6\right|=3,7\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1,6=3,7\\2x-1,6=-3,7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=5,3\\2x=-2,1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2,65\\x=-1,05\end{matrix}\right.\)
Vậy ....
b/ \(5,4-\left|3x-1,2\right|=5,5\)
\(\Leftrightarrow\left|3x-1,2\right|=-0,1\)
Mà \(\left|3x-1,2\right|\ge0\)
\(\Leftrightarrow x\in\varnothing\)
c/ \(\left|x+1,3\right|+\left|x+2,4\right|=4x\)
Mà \(\left\{{}\begin{matrix}\left|x+1,3\right|\ge0\\\left|x+2,4\right|\ge0\end{matrix}\right.\) \(\Leftrightarrow4x\ge0\)
\(\Leftrightarrow x+1,3+x+2,4=4x\)
\(\Leftrightarrow2x+3,7=4x\)
\(\Leftrightarrow3,7=4x-2x\)
\(\Leftrightarrow2x=3,7\)
\(\Leftrightarrow x=1,85\)
Vậy ....
d/ \(\left|x-1,2\right|+\left|2,5-x\right|=0\)
Mà \(\left\{{}\begin{matrix}\left|x-1,2\right|\ge0\\\left|2,5-x\right|\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left|x-1,2\right|=0\\\left|2,5-x\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1,2=0\\2,5-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1,2\\x=2,5\end{matrix}\right.\) (loại)
Vậy ..
a, \(\left|2x-1,6\right|-2,3=1,4\)
\(\Rightarrow\left|2x-1,6\right|=3,7\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1,6=3,7\\2x-1,6=-3,7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2,65\\x=-1,05\end{matrix}\right.\)
b,\(5,4-\left|3x-1,2\right|=5,5\)
\(\Rightarrow\left|3x-1,2\right|=-0,1\) (vô lí)
Vì \(\left|x\right|\ge0\) mà \(\left|3x-1,2\right|< 0\)
Vậy, không có giá trị của x thỏa mãn.
c, \(\left|x+1,3\right|+\left|x+2,4\right|=4x\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x+1,3\right|\ge0\\\left|x+2,4\right|\ge0\end{matrix}\right.\Leftrightarrow4x\ge0\)
\(\Leftrightarrow x+1,3+x+2,4=4x\)
\(\Leftrightarrow x+x+1,3+2,4=4x\)
\(\Leftrightarrow2x+3,7=4x\)
\(\Leftrightarrow2x-4x=-3,7\)
\(\Leftrightarrow-2x=-3,7\)
\(\Leftrightarrow x=\dfrac{3,7}{2}\)
d, \(\left|x-1,2\right|+\left|2,5-x\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x-1,2\right|\ge0\\\left|2,5-x\right|\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left|x-1,2\right|=0\\\left|2,5-x\right|=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1,2=0\\2,5-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1,2\\x=2,5\end{matrix}\right.\)
Tìm x \(\in\) Q biết:
a)\(\left|2,5-x\right|-1,3=0\)
b)\(1,6\cdot\left|x-0,2\right|=0\)
c)\(\left|\dfrac{1}{3}-x\right|-\left|\dfrac{-3}{7}\right|=0\)
d)\(\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)
e)\(\left|x-1,5\right|+\left|2,5-x\right|=0\)
a) \(\left|2,5-x\right|-1,3=0\)
th1: \(2,5-x\ge0\Leftrightarrow x\le2,5\)
\(\Rightarrow\left|2,5-x\right|-1,3=0\Leftrightarrow2,5-x-1,3=0\Leftrightarrow x=1,2\left(tmđk\right)\)
th2: \(2,5-x< 0\Leftrightarrow x>2,5\)
\(\Rightarrow\left|2,5-x\right|-1,3=0\Leftrightarrow x-2,5-1,3=0\Leftrightarrow x=3,8\left(tmđk\right)\)
vậy \(x=1,2;x=3,8\)
b) \(1,6.\left|x-0,2\right|=0\Leftrightarrow\left|x-0,2\right|=0\Leftrightarrow x-0,2=0\Leftrightarrow x=0,2\) vậy \(x=0,2\)
c) \(\left|\dfrac{1}{3}-x\right|-\left|\dfrac{-3}{7}\right|=0\)
th1: \(\dfrac{1}{3}-x\ge0\Leftrightarrow x\le\dfrac{1}{3}\)
\(\Rightarrow\left|\dfrac{1}{3}-x\right|-\left|\dfrac{-3}{7}\right|=0\Leftrightarrow\dfrac{1}{3}-x-\dfrac{3}{7}=0\Leftrightarrow x=\dfrac{-2}{21}\left(tmđk\right)\)
th2: \(\dfrac{1}{3}-x< 0\Leftrightarrow x>\dfrac{1}{3}\)
\(\Rightarrow\left|\dfrac{1}{3}-x\right|-\left|\dfrac{-3}{7}\right|=0\Leftrightarrow x-\dfrac{1}{3}-\dfrac{3}{7}=0\Leftrightarrow x=\dfrac{16}{21}\left(tmđk\right)\)
vậy \(x=\dfrac{-2}{21};x=\dfrac{16}{21}\)
d) \(\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)
th1: \(x+\dfrac{4}{15}\ge0\Leftrightarrow x\ge\dfrac{-4}{15}\)
\(\Rightarrow\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\Leftrightarrow x+\dfrac{4}{15}-3,75=-2,15\)
\(\Leftrightarrow x=\dfrac{4}{3}\left(tmđk\right)\)
th2: \(x+\dfrac{4}{15}< 0\Leftrightarrow x< \dfrac{-4}{15}\)
\(\Rightarrow\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\Leftrightarrow-x-\dfrac{4}{15}-3,75=-2,15\)
\(\Leftrightarrow x=\dfrac{-28}{15}\left(tmđk\right)\)
vậy \(x=\dfrac{4}{3};x=\dfrac{-28}{15}\)
e) ta có : \(\left|x-1,5\right|\ge0\forall x\) và \(\left|2,5-x\right|\ge0\forall x\)
\(\Rightarrow\left|x-1,5\right|+\left|2,5-x\right|=0\Leftrightarrow\left\{{}\begin{matrix}x-1,5=0\\2,5-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1,5\\x=2,5\end{matrix}\right.\) 2 giá trị này khác nhau \(\Rightarrow\) phương trình vô nghiệm
tìm x \(\in\) Q biết
\(a,\left|2,5-x\right|=1,3\)
\(b,1,6-\left|x-0,2\right|=0\)
\(c,\left|x-1,5\right|+\left|2,5-x\right|=0\)
\(d,\left(x-\frac{1}{2}\right)^2=0\)
\(e,\left(x-2\right)^2=1\)
\(f\left(2x-1\right)^3=-8\)
Tìm x:
a) \(3x\left(3x-8\right)-9x^2+8=0\)
b)\(6x-15-x\left(5-2x\right)=0\)
c) \(x^3-16x=0\)
d) \(2x^2+3x-5=0\)
e) \(3x^2-x\left(3x-6\right)=36\)
f) \(\left(x+2\right)^2-\left(x-5\right)\left(x+1\right)=17\)
g) \(\left(x-4\right)^2-x\left(x+6\right)=9\)
h) \(4x\left(x-1000\right)-x+1000=0\)
i) \(x^2-36=0\)
j) \(x^2y-2+x+x^2-2y+xy=0\)
k) \(x\left(x+1\right)-\left(x-1\right).\left(2x-3\right)=0\)
l) \(3x^3-27x=0\)
Tìm x biết :
a) \(\left(x-2\right)^3+6\left(x+1\right)^2-x^3+12=0\)
b) \(\left(x-5\right)\left(x+5\right)-\left(x+3\right)^3+3\left(x-2\right)^2=\left(x+1\right)^2-\left(x+4\right)\left(x-4\right)+3x^2\)
c) \(\left(2x+3\right)^2+\left(x-1\right)\left(x+1\right)=5\left(x+2\right)^2-\left(x-5\right)\left(x+1\right)+\left(x+4\right)^2\)
d) \(\left(1-3x\right)^2-\left(x-2\right)\left(9x+1\right)=\left(3x-4\right)\left(3x+4\right)-9\left(x+3\right)^2\)
a/ \(x=\dfrac{-5}{12}\)
b/ \(x\approx-1,9526\)
c/ \(x=\dfrac{21-i\sqrt{199}}{10}\)
d/ \(x=\dfrac{-20}{13}\)
a) (x-2)3+6(x+1)2-x3+12=0
⇒ x3-6x2+12x-8+6(x2+2x+1)-x3+12=0
⇒ x3-6x2+12x-8+6x2+12x+6-x3+12=0
⇒ 24x+10=0
⇒ 24x=-10
⇒ x=-5/12
a.
PT \(\Leftrightarrow x^3-6x^2+12x-8+6(x^2+2x+1)-x^3+12=0\)
\(\Leftrightarrow x^3-6x^2+12x-8+6x^2+12x+6-x^3+12=0\)
\(\Leftrightarrow 24x+10=0\Leftrightarrow x=\frac{-5}{12}\)
b. Bạn xem lại đề, nghiệm khá xấu không phù hợp với mức độ tổng thể của bài.
c.
PT $\Leftrightarrow (4x^2+12x+9)+(x^2-1)=5(x^2+4x+4)+(x^2-4x-5)+9(x^2+6x+9)$
$\Leftrightarrow 10x^2+42x+64=0$
$\Leftrightarrow x^2+(3x+7)^2=-15< 0$ (vô lý)
Do đó pt vô nghiệm.
d.
PT $\Leftrightarrow (1-6x+9x^2)-(9x^2-17x-2)=(9x^2-16)-9(x^2+6x+9)$
$\Leftrightarrow 11x+3=-54x-97$
$\Leftrightarrow 65x=-100$
$\Leftrightarrow x=\frac{-20}{13}$
Tìm \(x\in\mathbb{Q}\), biết :
a) \(\left|2,5-x\right|=1,3\)
b) \(1,6-\left|x-0,2\right|=0\)
c) \(\left|x-1,5\right|+\left|2,5-x\right|=0\)
a) Vì \(\left|2,5-x\right|=1,3\Rightarrow\left\{{}\begin{matrix}2,5-x=1,3\\2,5-x=-1,3\end{matrix}\right.\left\{{}\begin{matrix}x=1,2\\x=3,8\end{matrix}\right.\)
b) \(1,6-\left|x-0,2\right|=0\)
\(\Rightarrow\left|x-0,2\right|=1,6\)
Vì \(\left|x-0,2\right|=1,6\Rightarrow\left\{{}\begin{matrix}x-0,2=1,6\\x-0,2=-1,6\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1,8\\x=-1,4\end{matrix}\right.\)
c) Vì \(\left|x-1,5\right|\ge0;\left|2,5-x\right|\ge0\)
Mà \(\left|x-1,5\right|+\left|2,5-x\right|=0\left\{{}\begin{matrix}x-1,5=0\\2,5-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1,5\\x=2,5\end{matrix}\right.\)
Vô lý vì \(x\) không thể nhận đồng thời 2 giá trị \(\Rightarrow x\) không có giá trị thỏa mãn đề bài
a. Vì |2,5 – x| = 1,3 nên 2,5 – x =1,3
=> x = 2,5 – 1,3 => x = 1,2
Hoặc 2,5 – x = -1,3 => x = 2,5 – ( -1,3)
=> x = 2,5 + 1,3 => x = 3,8
Vậy x = 1,2 hoặc x = 3,8
b. 1,6 - | x – 0,2| = 0 => |x – 0,2 | =1,6 nên x – 0,2 – 1,6
=> x = 1,6 + 0,2 => x = 1,8
Hoặc x – 0,2 = -1,6 => x= -1,6 + 0,2 => x = -1,4
Vậy x = 1,8 hoặc x = -1,4
c. |x – 1,5 | + | 2,5 – x | = 0 nên |x – 1,5| ≥ 0 ; |2,5 – x| ≥ 0
Suy ra: x – 1,5 = 0; 2,5 – x = 0 => x= 1,5 và x = 2,5
Điều này không đồng thời xảy ra. Vậy không có giá trị nào của x thoả mãn bài toán.
Bằng cách phân tích vế trái thành nhân tử, giải các PT sau:
a) \(2x.\left(x-3\right)+5\left(x-3\right)\)
b) \(\left(x^2-4\right)+\left(x-2\right).\left(3-2x\right)=0\)
c) \(x^3-3x^2+3x-1=0\)
a: =(x-3)(2x+5)
b: \(\Leftrightarrow\left(x-2\right)\left(x+2+3-2x\right)=0\)
=>(x-2)(5-x)=0
=>x=2 hoặc x=5
c: =>x-1=0
hay x=1
Bài 1:
a) |2x - 3| - \(\dfrac{1}{3}\)= 0
b) \(\dfrac{5}{6}-\left|x+\dfrac{1}{4}\right|=\dfrac{1}{4}\)
c) \(\left|2x-1\right|-\left|x+\dfrac{1}{3}\right|=0\)
d) \(3x-\left|x+15\right|=\dfrac{5}{4}\)
Bài 2:
a) A= 1,3 + 2,5
b) B= -4,3 - 13,7 + (-5,7) - 6,3
c) C= 25.(-5).(-0,4).(-0,2)
d) D=|11,4 - 3.4| + |12,4 - 15,5|
a, \(\left|2x-3\right|-\dfrac{1}{3}=0\Leftrightarrow\left|2x-3\right|=\dfrac{1}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=\dfrac{1}{3}\\2x-3=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\)
b, tương tự
c, \(\left|2x-1\right|-\left|x+\dfrac{1}{3}\right|=0\Leftrightarrow\left|2x-1\right|=\left|x+\dfrac{1}{3}\right|\)
TH1 : \(2x-1=x+\dfrac{1}{3}\Leftrightarrow x=\dfrac{4}{3}\)
TH2 : \(2x-1=-x-\dfrac{1}{3}\Leftrightarrow3x=\dfrac{2}{3}\Leftrightarrow x=\dfrac{2}{9}\)
d, \(3x-\left|x+15\right|=\dfrac{5}{4}\Leftrightarrow\left|x+15\right|=3x-\dfrac{5}{4}\)ĐK : x >= 5/12
TH1 : \(x+15=3x-\dfrac{5}{4}\Leftrightarrow-2x=-\dfrac{65}{4}\Leftrightarrow x=\dfrac{65}{8}\)( tm )
TH2 : \(x+15=\dfrac{5}{3}-3x\Leftrightarrow4x=-\dfrac{40}{3}\Leftrightarrow x=-\dfrac{10}{3}\)
b: ta có: \(\dfrac{5}{6}-\left|x+\dfrac{1}{4}\right|=\dfrac{1}{4}\)
\(\Leftrightarrow\left|x+\dfrac{1}{4}\right|=\dfrac{7}{12}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{4}=\dfrac{7}{12}\\x+\dfrac{1}{4}=-\dfrac{7}{12}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{5}{6}\end{matrix}\right.\)