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Nhân 2 vế với x ta có :
\(\left(2m-1\right)x+5=x+1\)
\(< =>\left(2m-1\right)x^2+5x=x^2+x\)
\(< =>\left(2m-1\right)x^2-x^2+5x-x=0\)
\(< =>x^2\left(2m-2\right)+x\left(5-1\right)=0\)
\(< =>x\left[x\left(2m-2\right)+1\left(5-1\right)\right]=0\)
\(< =>x\left[2xm-2x+4\right]=0\)
\(< =>x\left[2\left(mx-x+2\right)\right]=0\)
\(< =>\orbr{\begin{cases}x=0\\2\left(mx-x+2\right)=0\end{cases}< =>\orbr{\begin{cases}x=0\\mx-x+2=0\end{cases}< =>x=0< =>m\in}}ℤ\)
Để hiệu hai số là 15 thì 2 số sẽ có hiệu là 5
Xét các trường hợp : 6-1 , 7-2 , 8-3 , 9-4 , 5-0
Mà tổng hai chữ số là 10
--> Không tồn tại số đó
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ABCHKIEF
a)
Xét \(\Delta\)ABC và \(\Delta\)HBA có:
^BAC = ^BHA ( = 90 độ )
^ABC = ^HBA ( ^B chung )
=> \(\Delta\)ABC ~ \(\Delta\)HBA
b) AB = 3cm ; AC = 4cm
Theo định lí pitago ta tính được BC = 5 cm
Từ (a) => \(\frac{AB}{BH}=\frac{BC}{AB}\Rightarrow BH=\frac{AB^2}{BC}=1,8\)m
c) Xét \(\Delta\)AHC và \(\Delta\)AKH có: ^AKH = ^AHC = 90 độ
và ^HAC = ^HAK ( ^A chung )
=> \(\Delta\)AHC ~ \(\Delta\)AKH
=> \(\frac{AH}{AK}=\frac{AC}{AH}\Rightarrow AH^2=AC.AK\)
d) Bạn kiểm tra lại đề nhé!
\(\frac{x}{25}-\frac{x}{30}=\frac{1}{2}\)
\(\Leftrightarrow\frac{6x}{150}-\frac{5x}{150}=\frac{75}{150}\)
\(\Leftrightarrow6x-5x=75\)
\(\Leftrightarrow x=75\)
Gọi chiều rộng ban đầu là x ( x> 0; m )
=> Chiều dài ban đầu là: 3x ( m )
Diện tích ban đầu là: x . 3x = 3x^2 ( m^2 )
Tăng chiều rộng lên 2m ta được: x + 2 ( m )
Giảm chiều dài đi 4 m ta được: 3x - 4 (m )
Diện tích mới là: ( x + 2 ) ( 3x - 4 ) m^2 '
Vì diện tích tăng thêm 28m^2 nên ta có phương trình:
3x^2 + 28 = ( x + 2 ) ( 3x - 4 )
Giải ra ta tìm được: x = 18 m
Vậy diện tích ban đầu của miếng đất là: 3.18^2 = 972 ( m^2)
Gọi chiều rộng miếng đất HCN là: x
Chiều dài miếng đất HCN là: 3x
Diện tích miếng đất là: x.3x = 3x2
Theo đề ra, ta có phương trình:
(x + 2)(3x - 4) = 3x2 + 28
<=> 3x2 + 2x - 8 = 3x2 + 28
<=> 3x2 - 3x2 + 2x = 28 + 8
<=> 2x = 36
<=> x = 18
Vậy chiều rộng mảnh đất là 18 m, chiều dài mảnh đất là 18.3 = 54 m
Diện tích mảnh đất ban đầu là: 18.54 = 972 m2
\(26=\frac{x+\left(x-5\right)}{2}\)
\(\Leftrightarrow x+\left(x-5\right)=26\cdot2\)
\(\Leftrightarrow x+x-5=52\)
\(\Leftrightarrow2x=57\)
\(\Leftrightarrow x=\frac{57}{2}\)
\(26=\frac{x+\left(x-5\right)}{2}\)
\(\Leftrightarrow\frac{52}{2}=\frac{x+x-5}{2}\)
\(\Leftrightarrow52=x+x-5\)
\(\Leftrightarrow2x-5=52\)
\(\Leftrightarrow2x=57\)
\(\Leftrightarrow x=\frac{57}{2}=28,5\)
