a)27^3-72x=0
b)x^2+4x+4=6(x+2)
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Những câu hỏi liên quan
a) x^2+4x+4=6*(x+2)
b)27^3-72x=0
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Nếu là bài tìm x thì mình xin làm như sau
a) Ta có: \(x^2+4x+4=6\left(x+2\right)\)
\(\Rightarrow\left(x+2\right)^2=6\left(x+2\right)\)
\(\Rightarrow\left(x+2\right)^2-6\left(x+2\right)=0\)
\(\Rightarrow\left(x+2\right)\left(x+2-6\right)=0\)
\(\Rightarrow\left(x+2\right)\left(x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+2=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=4\end{matrix}\right.\)
Vậy: \(x\in\left\{-2;4\right\}\)
b) ta có: \(27^3-72x=0\)
\(\Rightarrow19683-72x=0\)
hay \(72x=19683\)
hay x=\(\frac{19683}{72}=273,375\)
Vậy: \(x=273,375\)
Đây là bài tìm x hả bạn
a : (x-\(\dfrac{1}{2}\))^2=0
b: (x-2)^2=1
c: (2x-1)^3=-8
d: (x+\(\dfrac{1}{2}\))^2=\(\dfrac{1}{16}\)
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a) \(\left(x-\dfrac{1}{2}\right)^2=0\)
\(\Rightarrow x-\dfrac{1}{2}=0\)
\(\Rightarrow x=\dfrac{1}{2}\)
b) \(\left(x-2\right)^2=1\)
\(\Rightarrow x-2=1\)
\(\Rightarrow x=3\)
c) \(\left(2x-1\right)^3=-8\)
\(\Rightarrow\left(2x-1\right)^3=\left(-2\right)^3\)
\(\Rightarrow2x-1=-2\)
\(\Rightarrow2x=-1\)
\(\Rightarrow x=\dfrac{-1}{2}\)
d) \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
\(\Rightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{1}{4}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\\x+\dfrac{1}{2}=-\dfrac{1}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{4}\\x=\dfrac{-3}{4}\end{matrix}\right.\).
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Bình luận (3)
a , \(\left(x-\dfrac{1}{2}\right)^2=0\)
<=> \(x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\)
b , \(\left(x-2\right)^2=1\Rightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
c , \(\left(2x-1\right)^3=-8\Rightarrow2x-1=-2\Rightarrow x=\dfrac{-1}{2}\)
d , \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{4^2}\)
<=> \(\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\\x+\dfrac{1}{2}=\dfrac{-1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{4}\\x=\dfrac{-3}{4}\end{matrix}\right.\)
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Bình luận (1)
a) \(\left(x-\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0^2\)
\(\Leftrightarrow x-\dfrac{1}{2}=0\)
\(\Leftrightarrow x=0+\dfrac{1}{2}\)
\(\Leftrightarrow x=\dfrac{1}{2}\left(TM\right)\)
Vậy \(x=\dfrac{1}{2}\) là giá trị cần tìm
b) \(\left(x-2\right)^2=1\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)^2=1^2\\\left(x-2\right)^2=\left(-1\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=1\\x-2=\left(-1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=1\end{matrix}\right.\) \(\left(TM\right)\)
Vậy \(x\in\left\{3;1\right\}\)
c) \(\left(2x-1\right)^3=-8\)
\(\Rightarrow\left(2x-1\right)^3=\left(-2\right)^3\)
\(\Rightarrow2x-1=-2\)
\(\Rightarrow2x=-2+1\)
\(\Rightarrow2x=-1\)
\(\Rightarrow x=\dfrac{-1}{2}\left(TM\right)\)
Vậy \(x=\dfrac{-1}{2}\)
d) \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{4}\\\left(x+\dfrac{1}{2}\right)^2=\dfrac{-1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\\x+\dfrac{1}{2}=-\dfrac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-1}{4}\\x=\dfrac{-3}{4}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{-1}{4};\dfrac{-3}{4}\right\}\) là giá trị cần tìm
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Xem thêm câu trả lời
a)tim x
1/(3x-1)2-(x+7)2=0
giup mik nha cac ban
Ta có : \(\left(3x-1\right)^2-\left(x+7\right)^2=0\)
\(\left(3x-1+x+7\right)\left(3x-1-x-7\right)=0\)
\(\left(4x+6\right)\left(2x-8\right)=0\)
Nên : 4x + 6 = 0 hoặc 2x - 8 = 0
4x = -6 hoặc 2x = 8
x = \(\frac{-3}{2}\) hoặc x = 4
Vậy x = \(\frac{-3}{2}\) hoặc x = 4
Cac ban giup minh voi
1) Giai cac phuong trinh
a) 2010.(4x-3)-4x2+3=0
b)( x2-\(\frac{25}{4}\))2= 10x +1
1) tim 5 gia tri cua x
a) 4.(x-3)<0
b) -2.(x+1)<0
Mong cac ban giup do!!!! giup mik clik cho
a) 4.(x-3)<0 khi 4 và x-3 là hai số nguyên khác dấu
mà 4>0 suy ra x-3<0
x<3
Vậy với x<3 thì 4.(x-3)<0
b) -2.(x+1)<0 khi -2 và x+1 là hai số nguyên khác dấu
mà -2<0 suy ra x+1>0
x>1
Vậy với x>1 thì -2.(x+1)<0
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tim y
y-6:2-<48-24x2:6-3>=0
cac ban biet nho giup mik nha
y - 6 : 2 - ( 48 - 48 : 6 - 3 ) = 0
y - 6 : 2 - ( 48 - 8 - 3 ) = 0
y - 6 : 2 - 37 = 0
y - 3 - 37 = 0
y = 37 + 3
y = 40
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\(\left(x-\frac{3}{4}\right)^2+\left(x-\frac{3}{4}\right)\left(x-\frac{1}{2}\right)=0\)
\(\left(4x-\frac{4x-3}{5}-\frac{2\left(x+3\right)}{7}\right)=0\)
Giup minh voi cac ban oi
a) <=>(x - 3/4)(x-3/4 +x-1/2)=0
<=>(x-3/4)(2x-5/4)=0
<=>x-3/4=0 hoặc 2x-5/4=0
<=>x=3/4 hoặc x=5/8
Vậy tập nghiệm của phương trình trên là S={3/4;5/8}
b)<=>140x/35 - 7(4x-3)/35 - 10(x+3)/35=0
<=>140x-28x+21-10x-30=0
<=>102x=9
<=>x=3/34
Vậy tập nghiệm của phương trình trên là S={3/34}
1. tim xa) x+4 chia het x+1 b) x-7 chia het x-3 c) (4x+3) chia het x+2 d) 4x-5 chia het x2. tim x,ya) (x-3).(2y+1)7b) (2x+1).(3y-2)-553. tim xa) (x+1)+(x+3)+(x+5)+...+(x+99)0b) (x-3)+(x-2)+(x-1)+...+10+1111 CAC BAN GIUP MINH CLIK CHO, CAM ON CAC BAN RAT NHIEU !!! CHIEU MIK PHAI NOP ROI
Đọc tiếp
1. tim x
a) x+4 chia het x+1 b) x-7 chia het x-3 c) (4x+3) chia het x+2 d) 4x-5 chia het x
2. tim x,y
a) (x-3).(2y+1)=7
b) (2x+1).(3y-2)=-55
3. tim x
a) (x+1)+(x+3)+(x+5)+...+(x+99)=0
b) (x-3)+(x-2)+(x-1)+...+10+11=11
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(x-3) chia het cho (x+1) ; (2x+5) chia het cho (x+1); (4x+1)chia het cho (2x+2)
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