Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Câu 1 :
- Gọi chiều dài miếng đất là x ( m, x > 6 )
=> Chiều rộng miếng đất là : x - 6 ( m )
=> Chu vi miếng đất đó là : \(2\left(x+x-6\right)\) ( m )
Theo đề bài chu vi mảnh đất đó là 60m nên ta có phương trình :
\(2\left(x+x-6\right)=60\)
=> \(2x-6=30\)
=> \(2x=24\)
=> \(x=12\) ( TM )
Mà diện tích mảnh đất là : \(x\left(x-6\right)\)
=> Smảnh đất = \(12\left(12-6\right)=12.6=72\left(m^2\right)\)
Mình làm 2 câu ab thôi nhé!Cách giải các bài tập này đều như nhau!
Giải:
a) \(\frac{x-9}{x}-\frac{x}{x-9}=0\text{⇔}\frac{x-9}{x}=\frac{x}{x-9}\) (ĐKXĐ: x ≠ 0, x ≠ 9)
⇔ (x - 9)2 = x2 ⇔ (x - 9)2 - x2 = 0 ⇔ -9(2x + 9) = 0 ⇔ 2x + 9 = 0 ⇔ x = \(\frac{-9}{2}\)
Vậy phương trình trên có nghiệm là \(\frac{-9}{2}\)
b) \(\frac{x+3}{x-2}=\frac{5}{\left(x-2\right)\left(3-x\right)}\text{⇔}\frac{x+3}{5}=\frac{x-2}{\left(x-2\right)\left(3-x\right)}\text{⇔}\frac{x+3}{5}=\frac{1}{3-x}\) (ĐKXĐ: x ≠ 2, x ≠ 3)
⇔ (x + 3)(x - 3) = -5 ⇔ x2 - 9 = -5 ⇔ x2 = 4 ⇔ x = \(\pm\)2
Vậy phương trình có tập nghiêm S=\(\left\{\pm2\right\}\)
a, \(\frac{x-9}{x}-\frac{x}{x-9}=0\left(đkxđ:x\ne0;9\right)\)
\(< =>\frac{\left(x-9\right)^2}{x\left(x-9\right)}-\frac{x^2}{x\left(x-9\right)}=0\)
\(< =>x^2-18x+81-x^2=0\)
\(< =>18x=81< =>x=\frac{9}{2}\left(tmđk\right)\)
b) \(\frac{x-3}{x-2}+\frac{x+2}{x-4}=-1\)
\(\Rightarrow\frac{\left(x-3\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}+\frac{\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}=-1\)
\(\Rightarrow\frac{\left(x-3\right)\left(x-4\right)+x^2-4}{\left(x-2\right)\left(x-4\right)}=-1\)
\(\Rightarrow\frac{x^2-7x+12+x^2-4}{\left(x-2\right)\left(x-4\right)}=-1\)
\(\Rightarrow\frac{2x^2-7x+8}{\left(x-2\right)\left(x-4\right)}=-1\)
\(\Rightarrow\frac{2x^2-7x+8}{\left(x-2\right)\left(x-4\right)}=-1\)
.................
a) \(\frac{2}{x-1}+\frac{2x+3}{x^2+x+1}=\frac{\left(2x-1\right)\left(2x+1\right)}{x^3-1}\)
\(\Rightarrow\frac{2\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{\left(2x+3\right)\left(x-1\right)}{\left(x+1\right)\left(x^2+x+1\right)}=\frac{\left(2x-1\right)\left(2x+1\right)}{x^3-1}\)
\(\Rightarrow\frac{2\left(x^2+x+1\right)+\left(2x+3\right)\left(x-1\right)}{x^3-1}=\frac{\left(2x-1\right)\left(2x+1\right)}{x^3-1}\)
\(\Rightarrow\left(x^3-1\right)\left[2\left(x^2+x+1\right)+\left(2x+3\right)\left(x-1\right)\right]=\left(x^3-1\right)\left(2x-1\right)\left(2x+1\right)\)
\(\Rightarrow2\left(x^2+x+1\right)+\left(2x+3\right)\left(x-1\right)=\left(2x-1\right)\left(2x+1\right)\)
\(\Rightarrow2\left(x^2+x+1\right)+\left(2x+3\right)\left(x-1\right)-\left(2x-1\right)\left(2x+1\right)=0\)
\(\Rightarrow2x^2+2x+2+2x^2-2x+3x-3-\left(4x^2-1\right)=0\)
\(\Rightarrow2x^2+2x+2+2x^2-2x+3x-3-4x^2+1=0\)
\(\Rightarrow3x=0\)
\(\Rightarrow luon-dung-voi-moi-x\)
Bài 1:
a) Ta có: \(\frac{4}{5}x-3=\frac{1}{5}x\left(4x-15\right)\)
\(\Leftrightarrow\frac{4x}{5}-3=\frac{4x^2}{5}-3x\)
\(\Leftrightarrow\frac{12x}{15}-\frac{45}{15}-\frac{12x^2}{15}+\frac{45x}{15}=0\)
Suy ra: \(12x-45-12x^2+45x=0\)
\(\Leftrightarrow-12x^2+57x-45=0\)
\(\Leftrightarrow-12x^2+12x+45x-45=0\)
\(\Leftrightarrow-12x\left(x-1\right)+45\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(-12x+45\right)=0\)
\(\Leftrightarrow-3\left(x-1\right)\left(4x-15\right)=0\)
mà \(-3\ne0\)
nên \(\left[{}\begin{matrix}x-1=0\\4x-15=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\4x=15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{15}{4}\end{matrix}\right.\)
Vậy: Tập nghiệm \(S=\left\{1;\frac{15}{4}\right\}\)
b) Ta có: \(\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}=\frac{\left(x-3\right)\left(3-x\right)}{4}\)
\(\Leftrightarrow\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}+\frac{\left(x-3\right)^2}{4}=0\)
\(\Leftrightarrow\frac{12\left(x-3\right)}{12}-\frac{2\left(x-3\right)\left(2x-5\right)}{12}+\frac{3\left(x-3\right)^2}{12}=0\)
Suy ra: \(12\left(x-3\right)-2\left(2x^2-11x+15\right)+3\left(x^2-6x+9\right)=0\)
\(\Leftrightarrow12x-36-4x^2+22x-30+3x^2-18x+27=0\)
\(\Leftrightarrow-x^2+16x-39=0\)
\(\Leftrightarrow-\left(x^2-16x+39\right)=0\)
\(\Leftrightarrow x^2-13x-3x+39=0\)
\(\Leftrightarrow x\left(x-13\right)-3\left(x-13\right)=0\)
\(\Leftrightarrow\left(x-13\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-13=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\\x=3\end{matrix}\right.\)
Vậy: Tập nghiệm S={3;13}
c) Ta có: \(\frac{\left(3x+1\right)\left(3x-2\right)}{3}+5\left(3x+1\right)=\frac{2\left(2x+1\right)\left(3x+1\right)}{3}+2x\left(3x+1\right)\)
\(\Leftrightarrow\frac{9x^2-3x-2}{3}+5\left(3x+1\right)-\frac{12x^2+10x+2}{3}-2x\left(3x+1\right)=0\)
\(\Leftrightarrow\frac{9x^2-3x-2-12x^2-10x-2}{3}-6x^2+13x+5=0\)
\(\Leftrightarrow\frac{-3x^2-13x-4}{3}+\frac{3\left(-6x^2+13x+5\right)}{3}=0\)
Suy ra: \(-3x^2-13x-4-18x^2+39x+15=0\)
\(\Leftrightarrow-21x^2+26x+11=0\)
\(\Leftrightarrow-21x^2-7x+33x+11=0\)
\(\Leftrightarrow-7x\left(3x+1\right)+11\left(3x+1\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(-7x+11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\-7x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\-7x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{3}\\x=\frac{11}{7}\end{matrix}\right.\)
Vậy: Tập nghiệm \(S=\left\{-\frac{1}{3};\frac{11}{7}\right\}\)
\(a,\frac{1}{2}x+\frac{1}{2}+\frac{1}{4}x+\frac{3}{4}=3-\frac{1}{3}x-\frac{2}{3}\)
\(\frac{13}{12}x=\frac{13}{12}\Rightarrow x=1\)
a, \(\frac{x-3}{5}\) = 6 - \(\frac{1-2x}{3}\)
⇔ 3(x - 3) = 90 - 5(1 - 2x)
⇔ 3x - 9 = 90 - 5 + 10x
⇔ 3x - 10x = 90 - 5 + 9
⇔ -7x = 94
⇔ x = \(\frac{-94}{7}\)
S = { \(\frac{-94}{7}\) }
b, \(\frac{3x-2}{6}\) - 5 = \(\frac{3-2\left(x+7\right)}{4}\)
⇔ 2(3x - 2) - 60 = 9 - 6(x + 7)
⇔ 6x - 4 - 60 = 9 - 6x - 42
⇔ 6x + 6x = 9 - 42 + 60 + 4
⇔ 12x = 31
⇔ x = \(\frac{31}{12}\)
S = { \(\frac{31}{12}\) }
c, \(\frac{x+8}{6}\) - \(\frac{2x-5}{5}\) = \(\frac{x+1}{3}\) - x + 7
⇔ 5(x+ 8) - 6(2x - 5) = 10(x+1) - 30x+210
⇔ 5x+ 40 - 12x+ 30 = 10x+ 10 - 30x+210
⇔ 5x - 12x - 10x+ 30x = 10+ 210 - 30- 40
⇔ 13x = 150
⇔ x = \(\frac{150}{13}\)
S = { \(\frac{150}{13}\) }
d, \(\frac{7x}{8}\) - 5(x - 9) = \(\frac{2x+1,5}{6}\)
⇔ 21x - 120(x - 9) = 4(2x + 1,5)
⇔ 21x - 120x + 1080 = 8x + 6
⇔ 21x - 120x - 8x = 6 - 1080
⇔ -107x = -1074
⇔ x = \(\frac{1074}{107}\)
S = { \(\frac{1074}{107}\) }
e, \(\frac{5\left(x-1\right)+2}{6}\) - \(\frac{7x-1}{4}\) = \(\frac{2\left(2x+1\right)}{7}\) - 5
⇔ 140(x-1)+56 - 42(7x-1) = 48(2x+1)-840
⇔ 140x -140+56 -294x+42= 96x+48 -840
⇔ 140x -294x -96x = 48 -840 -42 -56+140
⇔ -250x = -750
⇔ x = 3
S = { 3 }
f, \(\frac{x+1}{3}\) + \(\frac{3\left(2x+1\right)}{4}\) = \(\frac{2x+3\left(x+1\right)}{6}\) + \(\frac{7+12x}{12}\)
⇔ 4(x+1)+9(2x+1) = 4x+6(x+1)+7+12x
⇔ 4x+4+18x+9 = 4x+6x+6+7+12x
⇔ 4x+18x - 4x - 6x - 12x = 6+7- 9 - 4
⇔ 0x = 0
S = R
Chúc bạn học tốt !
Bạn ơi giải giúp mình 2 bài này với ạ : https://hoc24.vn/hoi-dap/question/969683.html
Mình cảm ơn trước nhaa
a, Ta có: \(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x^2-2x}\)
\(\Leftrightarrow\frac{x+2}{x-2}-\frac{2}{x^2-2x}=\frac{1}{x}\)
\(Đkxđ:\left\{{}\begin{matrix}x\ne2\\x\ne0\end{matrix}\right.\)
\(Pt\Leftrightarrow x\left(x+2\right)-2=x-2\)
\(\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=-1\left(tmđk\right)\end{matrix}\right.\)
Vậy .........
\(b,Đkxđ:x\ne-5\)
Ta có: \(\frac{2x-5}{x+5}=3\)
\(\Leftrightarrow2x-5=3\left(x+5\right)\)
\(\Leftrightarrow x=20\left(tmđk\right)\)
Vậy .........
c, \(Đkxđ:x\ne3\)
Ta có: \(\frac{\left(x^2+2x\right)-\left(3x+6\right)}{x-3}=0\)
\(\Leftrightarrow x^2+2x-3x-6=0\)
\(\Leftrightarrow x^2-x-6=0\)
\(\Leftrightarrow x^2-3x+2x-6=0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\left(tm\right)\\x=3\left(ktmđk\right)\end{matrix}\right.\)
Vậy ............