a) |9+x|=2x b) |5x|-3x=2 c) |x+6|-9=2x d)|2x-3|+x=21
a) |9 + x| = 2x
b) |5x|-3x=2
c) |x+6| -9 =2x
d) |2x-3| + x =21
e) |4 + 2x| = -4x
i) |3x -1| +2 = x
g) |x+15| +1 = 3x
h) | 2x -5| +x =2
a) \(\left|x+9\right|=2x\)
\(\Leftrightarrow\left[{}\begin{matrix}x+9=2x\\x+9=-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-3\end{matrix}\right.\)
b) \(\left|5x\right|-3x=2\Leftrightarrow\left|5x\right|=3x+2\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=3x+2\\-5x=3x+2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{-1}{4}\end{matrix}\right.\)
c) \(\left|x+6\right|-9=2x\Leftrightarrow\left|x+6\right|=2x+9\)
\(\Leftrightarrow\left[{}\begin{matrix}x+6=2x+9\\-x-6=2x+9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)
d) \(\left|2x-3\right|+x=21\Leftrightarrow\left|2x-3\right|=21-x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=21-x\\2x-3=x-21\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-18\end{matrix}\right.\)
e) \(\left|2x+4\right|=-4x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+4=4x\\2x+4=-4x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{-2}{3}\end{matrix}\right.\)
i) \(\left|3x-1\right|+2=x\Leftrightarrow\left|3x-1\right|=x-2\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=x-2\\3x-1=2-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=\frac{3}{4}\end{matrix}\right.\)
g) \(\left|x+15\right|+1=3x\Leftrightarrow\left|x+15\right|=3x-1\)
\(\Leftrightarrow\left[{}\begin{matrix}x+15=3x-1\\x+15=1-3x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-3,5\end{matrix}\right.\)
h) \(\left|2x-5\right|+x=2\Leftrightarrow\left|2x-5\right|=2-x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=2-x\\2x-5=x-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{3}\\x=3\end{matrix}\right.\)
a) |9+x|=2x
TH1: 9+x=2x
<=> 9=2x-x
<=> x=9
TH2: -9-x=2x
<=> -9=3x
<=> x=-3
b) |5x|-3x=2
TH1: 5x-3x=2
<=> 2x=2
<=> x=1
TH2: -5x-3x=2
<=> -8x=2
<=>x=-4
c) |x+6|-9=2x
TH1: x+6-9=2x
<=> -3=x
TH2: -x-6-9=2x
<=> -15=3x
<=>x=-5
d) |2x-3|+x=21
TH1: 2x-3+x=21
<=> 3x=24
<=> x=8
TH2: -2x+3+x=21
<=> -x=18
<=> x=-18
e,i,g,h tương tự
Tìm x, biết
a, | 9 + x| = 2x
b, | 5x | - 3x = 2
c, | x + 6 | - 9 = 2x
d, | 2x – 3| + x = 21
a) \(\left|9+x\right|=2x\)
\(\Rightarrow\left[{}\begin{matrix}9+x=2x\\9+x=-2x\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-x=9\\-2x-x=9\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=9\\-3x=9\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=9\\x=-3\end{matrix}\right.\)
b) \(\left|5x\right|-3x=2\)
\(\Leftrightarrow\left|5x\right|=2+3x\)
\(\Rightarrow\left[{}\begin{matrix}5x=2+3x\\5x=-2-3x\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}5x-3x=2\\5x+3x=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=2\\8x=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{4}\end{matrix}\right.\)
c) \(\left|x+6\right|-9=2x\)
\(\Leftrightarrow\left|x+6\right|=2x+9\)
\(\Rightarrow\left[{}\begin{matrix}x+6=2x+9\\x+6=-2x-9\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-2x=9-6\\x+2x=-9-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}-3x=3\\3x=-15\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=-5\end{matrix}\right.\)
d) \(\left|2x-3\right|+x=21\)
\(\Leftrightarrow\left|2x-3\right|=21-x\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=21-x\\2x-3=-21+x\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x+x=21+3\\2x-x=-21+3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x=24\\x=-18\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-18\end{matrix}\right.\)
Giair các phương trình sau
\(a,\left|5x\right|=x+2\) \(b,\left|7x-3\right|-2x+6=0\)
\(c,\left|2x-3\right|-21=x\) \(d,\left|9-x\right|=2x\)
\(e,\left|x-15\right|+1=3x\) \(f,\left|5-4x\right|=4-5x\)
Ai giúp mik với ạ mik đang cần gấp
Mấy ý này bản chất ko khác nhau nhé, mình làm mẫu, bạn làm tương tự mấy ý kia nhé
a, \(\left|5x\right|=x+2\)
Với \(x\ge0\)thì \(5x=x+2\Leftrightarrow x=\dfrac{1}{2}\)
Với \(x< 0\)thì \(5x=-x-2\Leftrightarrow6x=-2\Leftrightarrow x=-\dfrac{1}{3}\)
b, \(\left|7x-3\right|-2x+6=0\Leftrightarrow\left|7x-3\right|=2x-6\)
Với \(x\ge\dfrac{3}{7}\)thì \(7x-3=2x-6\Leftrightarrow5x=-3\Leftrightarrow x=-\dfrac{3}{5}\)( ktm )
Với \(x< \dfrac{3}{7}\)thì \(7x-3=-2x+6\Leftrightarrow9x=9\Leftrightarrow x=1\)( ktm )
Vậy phương trình vô nghiệm
giải các pt có giá trị tuyệt đối sau
a, /9+x/=2x
b, /x+6/=2x+9
c, /2x-3/=2x-3
d, /4+2x/=-4x
e, /5x/=3x-2
g,/-2,5x/=x-12
h, /5x/-3x-2=0
i,/-2x/+x-5x-3=0
k,/3-x/ =x2-x (x+4)=0
m, (x-1)2+/x+21/-x2-13=0
a,|9+x|=2x
b,|x+6|-9=2x
c,|2x-3|+x =21
d,|4+2x|=-4x
e,|5x-4|=|x+2|
f,|2x-6|+|x+3|=8
g,|x-2|+|x-5|=3
h,|x+5|+|x-3|=9
a,\(\left|9+x\right|=2x\)
\(\Leftrightarrow\left[{}\begin{matrix}9+x=2x\\9x+x=-2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}9=x\\9=-3x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-3\end{matrix}\right.\)
Vậy...
Trường hợp 2 chưa chắc chắn lắm!!!
a) \(\left|9+x\right|=2x\)
Xét trường hợp 1:
\(9+x=2x\)
\(\Leftrightarrow9+x-2x=0\)
\(\Leftrightarrow9-x=0\)
\(\Leftrightarrow x=9\)
Xét trường hợp 2:
\(9+x=-2x\)
\(\Leftrightarrow9+x-\left(-2x\right)=0\)
\(\Leftrightarrow9+x+2x=0\)
\(\Leftrightarrow9+3x=0\)
\(\Leftrightarrow3x=-9\)
\(\Leftrightarrow x=-9:3\)
\(\Leftrightarrow x=-3\)
Vậy x=9 hoặc x=-3
b) \(\left|x+6\right|-9=2x\)
\(\Leftrightarrow\left|x+6\right|=2x+9\)
Xét trường hợp 1:
\(x+6=2x+9\)
\(\Leftrightarrow x+6-\left(2x+9\right)=0\)
\(\Leftrightarrow x+6-2x-9=0\)
\(\Leftrightarrow-3-x=0\)
\(\Leftrightarrow x=-3\)
Xét trường hợp 2:
\(x+6=-\left(2x+9\right)\)
\(\Leftrightarrow x+6-\left[-\left(2x+9\right)\right]=0\)
\(\Leftrightarrow x+6+\left(2x+9\right)=0\)
\(\Leftrightarrow x+6+2x+9=0\)
\(\Leftrightarrow3x+15=0\)
\(\Leftrightarrow3x=-15\)
\(\Leftrightarrow x=-15:3\)
\(\Leftrightarrow x=-5\)
Vậy x=-3 hoặc x=-5
ở câu b,bạn Nguyễn Nam phải thêm điều kiện cho mỗi trường hợp
TH1: phải đi với ĐK x phải lớn hơn hoặc bằng -6
TH2:phải đi với Đk x phải bé hơn -6 nên kết quả x=-5(KTM)
Vậy x=-3
TÌM X
a) |5x|-3x=2
b) |x+6|-9=2x
c) |2x-3|+x=21
d) | 4+2x|= -4x
e) |3x-1|+2=x
f) |x+15|+1=3x
CÁC BẠN ƠI GIÚP MÌNH VỚI
Bn gửi từng chút một thôi, nhiều quá lười đọc !
BẠN ƠI GIÚP MÌNH NHANH ĐƯỢC KO MÌNH ĐANG CẦN GẤP
Bài 2:
a,x-2=-6-5x-16
b,-5x-[-3]-x=13-x
c,15-[x-7]=-21-x
d,3x+17=2+3x+15
e,x-45-[x-9]=-35-x
f,[-5]+x=15-x-2x
g,2x-[-17]=15
a)x-2=6-5x-16
6x= -8
x= -4/3
b)-5x-(-3)-x=13-x
-5x+3-x=13-x
-5x=10
x=-2
c)15-(x-7)=-21-x
15-x+7=-21-x
22=-21 ( vô lí)
x không có giá trị
D)3x+17=2+3x+15
17=17
x có vô số giá trị
e) x-45-(x-9)=-35-x
x-45-x+9=-35-x
-x=1
x=-1
f) -5+x=15-x-2x
4x=20
x=5
g) 2x-(-17) =15
2x+17=15
2x=-2
x=-1
Giải các phương trình chứa dấu giá trị tuyệt đối
a) |9+x|=2x
b)|x+6|=2x+9
c)|2x-3|=2x-3
d)|4+2x|=-4x
e)|5x|=3x-2
g)|-2,5x|=x-12
i)|5x|-3x-2=0
k)|-2x|+x-5x-3=0
Tìm x,tính nhanh:
a.\(2x^2+5x+8+\sqrt{x}=x^2+3x+35+x^2+2x-7.\)
b.\(3\sqrt{x}+7x+5=\sqrt{x}+4x-6+3x+18\)
c.\(8\sqrt{x}+2x-9=5x+7+6\sqrt{x}-3x-12\)
d.\(2\sqrt{3x}+11x-18=5x+3+6\sqrt{3x}+6x-21\)
a.\(2x^2+5x+8+\sqrt{x}=x^2+3x+35+x^2+2x-7\)
\(=2x^2+5x+8+\sqrt{x}=2x^2+5x+28\Leftrightarrow\sqrt{x}=20\Leftrightarrow x=400.\)
b.\(3\sqrt{x}+7x+5=\sqrt{x}+4x-6+3x+18\)
\(=3\sqrt{x}+7x+5=\sqrt{x}+7x+12\Leftrightarrow2\sqrt{x}=7\Leftrightarrow x=\frac{49}{4}.\)
c.\(8\sqrt{x}+2x-9=5x+7+6\sqrt{x}-3x-12.\)
\(=8\sqrt{x}+2x-9=2x+6\sqrt{x}-5\Leftrightarrow2\sqrt{x}=4\Leftrightarrow x=4.\)
d.\(2\sqrt{3x}+11x-18=5x+3+6\sqrt{3x}+6x-21\)
\(=2\sqrt{3x}+11x-18=11x+6\sqrt{3x}-19\Leftrightarrow4\sqrt{3x}=1\)
\(\Leftrightarrow\sqrt{3x}=\frac{1}{4}\Leftrightarrow3x=\frac{1}{16}\Leftrightarrow x=\frac{1}{48}.\)
a) \(2x^2+5x+8+\sqrt{x}=x^2+3x+35+x^2+2x-7\)
<=> \(2x^2+5x+8+\sqrt{x}=2x^2+5x+28\)
<=> \(2x^2+5x+8+\sqrt{x}-\left(2x^2+5\right)=28\)
<=> \(\sqrt{x}+8=28\)
<=> \(\sqrt{x}=28-8\)
<=> \(\sqrt{x}=20\)
<=> \(\left(\sqrt{x}\right)^2=20^2\)
<=> x = 400
=> x = 400
b) \(3\sqrt{x}+7x+5=\sqrt{x}+4x-6+3x+18\)
<=> \(3\sqrt{x}+7x+5=7x+\sqrt{x}+12\)
<=> \(3\sqrt{x}+5=7x+\sqrt{x}+12-7x\)
<=> \(3\sqrt{x}+5=\sqrt{x}+12\)
<=> \(3\sqrt{x}=\sqrt{x}+12-5\)
<=> \(3\sqrt{x}=\sqrt{x}+7\)
<=> \(3\sqrt{x}-\sqrt{x}=7\)
<=> \(2\sqrt{x}=7\)
<=> \(\sqrt{x}=\frac{7}{2}\)
<=> \(\left(\sqrt{x}\right)^2=\left(\frac{7}{2}\right)^2\)
<=> \(x=\frac{49}{4}\)
=> \(x=\frac{49}{4}\)
c) \(8\sqrt{x}+2x-9=5x+7+6\sqrt{x}-3x-12\)
<=> \(8\sqrt{x}+2x-9=2x+6\sqrt{x}-5\)
<=> \(8\sqrt{x}-9=2x+6\sqrt{x}-5-2x\)
<=> \(8\sqrt{x}-9=6\sqrt{x}-5\)
<=> \(8\sqrt{x}=6\sqrt{x}-5+9\)
<=> \(8\sqrt{x}=6\sqrt{x}+4\)
<=> \(8\sqrt{x}-6\sqrt{x}=4\)
<=> \(2\sqrt{x}=4\)
<=> \(\sqrt{x}=2\)
<=> \(\left(\sqrt{x}\right)^2=2^2\)
<=> x = 4
=> x = 4
d) \(2\sqrt{3x}+11x-18=5x+3+6\sqrt{3x}+6x-21\)
<=> \(2\sqrt{3x}+11x-18=11x+6\sqrt{3x}-18\)
<=> \(2\sqrt{3x}+11x-18-\left(11x-18\right)=6\sqrt{3x}\)
<=>\(2\sqrt{3x}=6\sqrt{3x}\)
<=> \(2\sqrt{3x}-6\sqrt{3x}=0\)
<=>\(-4\sqrt{3x}=0\)
<=> \(\sqrt{3x}=0\)
<=> \(\left(\sqrt{3x}\right)^2=0^2\)
<=> 3x = 0
<=> x = 0
=> x = 0