Đừng lạm dụng mạng quá bạn ạ, mình không có ý gì, chỉ mong bạn suy nghĩ thôi, hi !
đề ko phải là khó ! mỗi bài mk lm tầm 1 câu nha
\(1,4\left(3x-2\right)-3\left(x-4\right)=7x+20\)
\(12x-12-3x+12=7x+20\)
\(12x-12-3x+12-7x-20=0\)
\(2x-20=0\)
\(x=10\)
\(1,\left(4x-3\right)\left(2x-1\right)=\left(x-3\right)\left(4x-3\right)\)
\(8x^2-10x+3=4x^2-15x+9\)
\(8x^2-10x+3-4x^2+15x-9=0\)
\(4x^2+5x-6=0\)
\(4x\left(x+2\right)-3\left(x+2\right)=0\)
\(\left(4x-3\right)\left(x+2\right)=0\)
\(\left[{}\begin{matrix}4x=3\\x=-2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\frac{3}{4}\\x=-2\end{matrix}\right.\)
\(1,6x^2-5x+3=2x-3x\left(3-2x\right)\)
\(6x^2-5x+3=2x-9x+6x^2\)
\(6x^2-5x+3-2x+9x-6x^2=0\)
\(2x+3=0\)
\(x=-\frac{3}{2}\)
Bài 1:
1, 4(3x - 2) - 3(x - 4) = 7x + 20
\(\Leftrightarrow\) 12x - 8 - 3x + 12 = 7x + 20
\(\Leftrightarrow\) 9x + 4 = 7x + 20
\(\Leftrightarrow\) 9x - 7x = 20 - 4
\(\Leftrightarrow\) 2x = 16
\(\Leftrightarrow\) x = 8
Vậy S = {8}
2, (3x - 1)(x + 3) = (2 - x)(5 - 3x)
\(\Leftrightarrow\) 3x2 + 9x - x - 3 = 10 - 6x - 5x + 3x2
\(\Leftrightarrow\) 3x2 + 8x - 3 = 3x2 - 11x + 10
\(\Leftrightarrow\) 3x2 - 3x2 + 8x + 11x = 10 + 3
\(\Leftrightarrow\) 19x = 13
\(\Leftrightarrow\) x = \(\frac{13}{19}\)
Vậy S = {\(\frac{13}{19}\)}
Tui làm 2 bài 1 đã, chứ nhiều quá :))
Bài 1:
5, (x + 3)2 - (x - 3)2 = 6x + 18
\(\Leftrightarrow\) (x + 3 - x + 3)(x + 3 + x - 3) = 6x + 18
\(\Leftrightarrow\) 6 . 2x = 6x + 18
\(\Leftrightarrow\) 12x = 6x + 18
\(\Leftrightarrow\) 12x - 6x = 18
\(\Leftrightarrow\) 6x = 18
\(\Leftrightarrow\) x = 3
Vậy S = {3}
6, \(\frac{x+1}{35}+\frac{x+3}{33}=\frac{x+5}{31}+\frac{x+7}{29}\)
\(\Leftrightarrow\) \(\left(\frac{x+1}{35}+1\right)+\left(\frac{x+3}{33}+1\right)=\left(\frac{x+5}{31}+1\right)+\left(\frac{x+7}{29}+1\right)\)
\(\Leftrightarrow\) \(\frac{x+36}{35}+\frac{x+36}{33}=\frac{x+36}{31}+\frac{x+36}{29}\)
\(\Leftrightarrow\) \(\frac{x+36}{35}+\frac{x+36}{33}-\frac{x+36}{31}-\frac{x+36}{29}=0\)
\(\Leftrightarrow\) (x + 36)(\(\frac{1}{35}+\frac{1}{33}-\frac{1}{31}-\frac{1}{29}\)) = 0
\(\Leftrightarrow\) x + 36 = 0
\(\Leftrightarrow\) x = -36
Vậy S = {-36}
Hết bài 1 rồi nhá!!
Bài 1:
3, (x + 1)(2x - 3) - 3(x - 2) = 2(x - 1)2
\(\Leftrightarrow\) 2x2 - 3x + 2x - 3 - 3x + 6 = 2x2 - 4x + 2
\(\Leftrightarrow\) 2x2 - 4x + 3 = 2x2 - 4x + 2
\(\Leftrightarrow\) 2x2 - 2x2 - 4x + 4x = 2 - 3
\(\Leftrightarrow\) 0x = -1
\(\Rightarrow\) Phương trình vô nghiệm
Vậy S = {\(\varnothing\)}
4, (x - 1)3 - x(x + 1)2 = 5x(2 - x) - 11(x + 2)
\(\Leftrightarrow\) x3 - 3x2 + 3x - 1 - x3 - 2x2 - x = 10x - 5x2 - 11x - 22
\(\Leftrightarrow\) -5x2 + 2x - 1 = -5x2 - x - 22
\(\Leftrightarrow\) -5x2 + 5x2 + 2x + x = -22 + 1
\(\Leftrightarrow\) 3x = -21
\(\Leftrightarrow\) x = -7
Vậy S = {-7}
5) (x + 3)2 - (x - 3)2 = 6x + 18
<=> (x + 3 + x -3)(x + 3 - x + 3) - 6x - 18 = 0
<=> 2x + 6 - 6x - 18 = 0
<=> -4x -12 = 0
<=> -4(x + 3) = 0
<=> x + 3 = 0
<=> x = -3
Bài 2:
1) (4x - 3)(2x - 1) = (x - 3)(4x - 3)
<=> (4x - 3)(2x - 1) - (x - 3)(4x - 3) = 0
<=> (4x - 3)(2x - 1 - x + 3) = 0
<=> (4x - 3)(x + 2) = 0
<=> \(\left[{}\begin{matrix}4x-3=0\\x+2=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=\frac{3}{4}\\x=-2\end{matrix}\right.\)
Vậy \(S=\left\{\frac{3}{4};-2\right\}\)
2) 25x2 - 9 = (5x + 3)(2x + 1)
<=> (5x - 3)(5x + 3) - (5x + 3)(2x + 1) = 0
<=> (5x + 3)(5x - 3 - 2x - 1)= 0
<=> (5x + 3)(3x - 4) = 0
<=> \(\left[{}\begin{matrix}5x+3=0\\3x-4=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=-\frac{3}{5}\\x=\frac{4}{3}\end{matrix}\right.\)
Vậy \(S=\left\{-\frac{3}{5};\frac{4}{3}\right\}\)
Bài 2:
Câu 1 với 2 Trang làm rùi nhé
3, (3x - 4)2 - 4(x + 1)2 = 0
\(\Leftrightarrow\) 9x2 - 24x + 16 - 4x2 - 8x - 4 = 0
\(\Leftrightarrow\) (3x - 4)2 - (2x + 2)2 = 0
\(\Leftrightarrow\) (3x - 4 - 2x - 2)(3x - 4 + 2x + 2) = 0
\(\Leftrightarrow\) (x - 6)(5x - 2) = 0
\(\Leftrightarrow\) x - 6 = 0 hoặc 5x - 2 = 0
\(\Leftrightarrow\) x = 6 hoặc x = \(\frac{2}{5}\)
Vậy S = {6; \(\frac{2}{5}\)}
4, Hình như đề lỗi rồi thì phải, theo mk là như này:
2x2 + 7x2 + 7x + 2x = 0
\(\Leftrightarrow\) 2x(x + 1) + 7x(x + 1) = 0
\(\Leftrightarrow\) (x + 1)(2x + 7x) = 0
\(\Leftrightarrow\) 9x(x + 1) = 0
\(\Leftrightarrow\) 9x = 0 hoặc x + 1 = 0
\(\Leftrightarrow\) x = 0 hoặc x = -1
Vậy S = {0; -1}
Bài 2:
5, (x - 2)(x + 2)(x2 - 10) = 72
\(\Leftrightarrow\) (x2 - 4)(x2 - 10) = 72
\(\Leftrightarrow\) x4 - 10x2 - 4x2 + 40 = 72
\(\Leftrightarrow\) x4 - 14x2 + 40 - 72 = 0
\(\Leftrightarrow\) x4 - 14x2 - 32 = 0
\(\Leftrightarrow\) x4 - 14x2 + 49 - 81 = 0
\(\Leftrightarrow\) (x2 - 7)2 - 81 = 0
\(\Leftrightarrow\) (x2 - 7 - 9)(x2 - 7 + 9) = 0
\(\Leftrightarrow\) (x2 - 16)(x2 + 2) = 0
\(\Leftrightarrow\) (x - 4)(x + 4)(x2 + 2) = 0
Vì x2 \(\ge\) 0 với mọi x nên x2 \(\ge\) 2 > 0 với mọi x
\(\Rightarrow\) (x - 4)(x + 4) = 0
\(\Leftrightarrow\) x - 4 = 0 hoặc x + 4 = 0
\(\Leftrightarrow\) x = 4 và x = -4
Vậy S = {4; -4}
Gửi tạm câu 5 bài 2 đã câu 6 để tui nghiên cứu đã :))
Bài 2: Sau khi nghiên cứu xong :))
6, x4 + 2x3 - 3x2 - 8x - 4 = 0
\(\Leftrightarrow\) x4 + 2x3 + x2 - 4x2 - 8x - 4 = 0
\(\Leftrightarrow\) x2(x2 + 2x + 1) - 4(x2 + 2x + 1) = 0
\(\Leftrightarrow\) (x2 + 2x + 1)(x2 - 4) = 0
\(\Leftrightarrow\) (x + 1)2(x - 2)(x + 2) = 0
\(\Leftrightarrow\) x + 1 = 0 hoặc x - 2 = 0 hoặc x + 2 = 0
\(\Leftrightarrow\) x = -1 và x = 2 và x = -2
Vậy S = {-1; 2; -2}
Xong Bài 2 nhá :))
Bài 3: bài 1, 2, 3 Trang làm nhá :))
4, \(\frac{6x+5}{2}-\frac{10x+3}{4}=2x+\frac{2x+1}{2}\)
\(\Leftrightarrow\) \(\frac{2\left(6x+5\right)}{4}-\frac{10x+3}{4}=\frac{8x}{4}+\frac{2\left(2x+1\right)}{4}\)
\(\Leftrightarrow\) 2(6x + 5) - 10x - 3 = 8x + 2(2x + 1)
\(\Leftrightarrow\) 12x + 10 - 10x - 3 = 8x + 4x + 2
\(\Leftrightarrow\) 2x + 7 = 12x + 2
\(\Leftrightarrow\) 2x - 12x = 2 - 7
\(\Leftrightarrow\) -10x = -5
\(\Leftrightarrow\) x = 2
Vậy S = {2}
5, (x - 4)(x + 4) - 2(3x - 2) = (x - 4)2
\(\Leftrightarrow\) x2 - 16 - 6x + 4 - x2 + 8x - 16 = 0
\(\Leftrightarrow\) 2x - 28 = 0
\(\Leftrightarrow\) x = 14
Vậy S = {14}
6, (x + 1)3 - (x - 1)3 = 6(x2 + x + 1)
\(\Leftrightarrow\) (x + 1 - x + 1)[(x + 1)2 + (x + 1)(x - 1) + (x - 1)2] = 6x2 + 6x + 6
\(\Leftrightarrow\) 2(x2 + 2x + 1 + x2 - x + x - 1 + x2 - 2x + 1) = 6x2 + 6x + 6
\(\Leftrightarrow\) 2(3x2 + 1) = 6x2 + 6x + 6
\(\Leftrightarrow\) 6x2 + 2 = 6x2 + 6x + 6
\(\Leftrightarrow\) 6x2 + 2 - 6x2 - 6x - 6 = 0
\(\Leftrightarrow\) -6x - 4 = 0
\(\Leftrightarrow\) x = \(\frac{-2}{3}\)
Vậy S = {\(\frac{-2}{3}\)}
Chúc bạn học tốt!!
Bài 3:
1) 6x2 - 5x + 3 = 2x - 3x(3 - 2x)
<=> 6x2 - 5x + 3 - 2x + 3x(3 - 2x) = 0
<=> 6x2 - 5x + 3 - 2x + 9x - 6x2 = 0
<=> 2x + 3 = 0
<=> \(x=-\frac{3}{2}\)
Vậy \(S=\left\{-\frac{3}{2}\right\}\)
2) \(\frac{2\left(x-4\right)}{4}-\frac{3+2x}{10}=x+\frac{1-x}{5}\)
\(\Leftrightarrow\frac{10\left(x-4\right)}{20}-\frac{2\left(3+2x\right)}{20}-\frac{20x}{20}-\frac{4\left(1-x\right)}{20}=0\)
\(\Leftrightarrow10x-40-6-4x-20x-4+4x=0\)
\(\Leftrightarrow-10x-50=0\)
\(\Leftrightarrow10x=-50\Leftrightarrow x=-5\)
Vậy \(S=\left\{-5\right\}\)
3) \(\frac{2x}{3}+\frac{3x-5}{4}=\frac{3\left(2x-1\right)}{2}-\frac{7}{6}\)
\(\Leftrightarrow\frac{8x}{12}+\frac{3\left(3x-5\right)}{12}-\frac{18\left(2x-1\right)}{12}+\frac{14}{12}=0\)
\(\Leftrightarrow8x+9x-15-36x+18+14=0\)
\(\Leftrightarrow17-19x=0\)
\(\Leftrightarrow x=\frac{17}{19}\)
Vậy \(S=\left\{\frac{17}{19}\right\}\)
Bài 1.
1)
\(4\cdot\left(3x-2\right)-3\cdot\left(x-4\right)=7x+20\\ \Leftrightarrow12x-8-3x+12-7x-20=0\\ \Leftrightarrow2x-16=0\\ \Rightarrow x=\frac{16}{2}=8\)
2)
\(\left(3x-1\right)\cdot\left(x+3\right)=\left(2-x\right)\cdot\left(5-3x\right)\\ \Leftrightarrow3x^2+9x-x-3-10+6x+5x-3x^2=0\\ \Leftrightarrow19x-13=0\\ \Rightarrow x=\frac{13}{19}\)
3)
\(\left(x+1\right)\cdot\left(2x-3\right)-3\cdot\left(x-2\right)=2\cdot\left(x-1\right)^2\\ \Leftrightarrow2x^2-3x+2x-3-3x+6-2x^2+4x-2=0\\ \Leftrightarrow1=0\\ \Rightarrow x=\varnothing\)
4)
\(\left(x-1\right)^3-x\cdot\left(x+1\right)^2=5x\cdot\left(2-x\right)-xx\cdot\left(x+2\right)\\ \Leftrightarrow x^3-5x^2+3x-1-x^3-2x^2-x-10x+5x^2+11x+22=0\\ \Leftrightarrow3x+21=0\\ \Rightarrow x=-\frac{21}{3}=-7\)
5)
\(\left(x+3\right)^2-\left(x-3\right)^2=6x+18\\ \Leftrightarrow x^2+6x+9-x^2+6x-9-6x-18=0\\ \Leftrightarrow6x-18=0\\ \Rightarrow x=\frac{18}{6}=3\)
6)
\(\frac{x+1}{35}+\frac{x+3}{33}=\frac{x+5}{31}+\frac{x+7}{29}\\ \Leftrightarrow\frac{x+1}{35}+1+\frac{x+3}{33}+1=\frac{x+5}{31}+1+\frac{x+7}{29}+1\\ \Leftrightarrow\frac{x+36}{35}+\frac{x+36}{33}=\frac{x+36}{31}+\frac{x+36}{29}\\ \Leftrightarrow\frac{x+36}{35}+\frac{x+36}{33}-\frac{x+36}{31}-\frac{x+36}{29}=0\\ \Leftrightarrow\left(x+36\right)\cdot\left(\frac{1}{35}+\frac{1}{33}-\frac{1}{31}-\frac{1}{29}\right)=0\)
Do:
\(\frac{1}{35}+\frac{1}{33}-\frac{1}{31}-\frac{1}{29}\ne0\\ \Rightarrow x+36=0\\ \Rightarrow x=-36\)
Bài 2.
1)
\(\left(4x-3\right)\cdot\left(2x-1\right)=\left(x-3\right)\cdot\left(4x-3\right)\\ \Leftrightarrow\left(4x-3\right)\cdot\left(2x-1\right)-\left(x-3\right)\cdot\left(4x-3\right)=0\\ \Leftrightarrow\left(4x-3\right)\cdot\left(2x-1-x+3\right)=0\\ \Leftrightarrow\left(4x-3\right)\cdot\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}4x-3=0\\x+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{3}{4}\\x=-2\end{matrix}\right.\)
2)
\(25x^2-9=\left(5x+3\right)\cdot\left(2x+1\right)\\ \Leftrightarrow\left(5x-3\right)\cdot\left(5x+3\right)-\left(5x+3\right)\cdot\left(2x+1\right)=0\\ \Leftrightarrow\left(5x+3\right)\cdot\left(5x-3-2x-1\right)=0\\ \Leftrightarrow\left(5x+3\right)\cdot\left(3x-4\right)\\ \Rightarrow\left[{}\begin{matrix}5x+3=0\\3x-4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\frac{3}{5}\\x=\frac{4}{3}\end{matrix}\right.\)
3)
\(\left(3x-4\right)^2-4\cdot\left(x+1\right)^2=0\\ \Leftrightarrow9x^2-24x+16-4x^2-8x-4=0\\ \Leftrightarrow5x^2-32x+12=0\\ \Leftrightarrow12-2x-30x+5x^2=0\\ \Leftrightarrow2\cdot\left(6-x\right)-5x\cdot\left(6-x\right)=0\\ \Leftrightarrow\left(2-5x\right)\cdot\left(6-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}2-5x=0\\6-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{2}{5}\\x=6\end{matrix}\right.\)
4)
\(2x^3+7x^2+7x+2=0\\ \Leftrightarrow2\left(x^3+1\right)+7x\left(x+1\right)=0\\ \Leftrightarrow2\cdot\left(x+1\right)\cdot\left(x^2-x+1\right)+7x\cdot\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left[2\cdot\left(x^2-x+1\right)+7x\right]=0\\ \Leftrightarrow\left(x+1\right)\cdot\left(x^2+5x+2\right)=0\\ \Leftrightarrow\left(x+1\right)\cdot\left(2+4x+x+2x^2\right)=0\\ \Leftrightarrow\left(x+1\right)\cdot\left[2\cdot\left(1+2x\right)+x\cdot\left(1+2x\right)\right]=0\\ \Leftrightarrow\left(x+1\right)\cdot\left(2+x\right)\cdot\left(1+2x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\2+x=0\\1+2x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\\x=-2\\x=-\frac{1}{2}\end{matrix}\right.\)
5)
\(\left(x-2\right)\cdot\left(x+2\right)\left(x^2-10\right)=72\\ \Leftrightarrow\left(x^2-4\right)\cdot\left(x^2-10\right)-72=0\\\Leftrightarrow x^4-14x^2-32=0\\ \Leftrightarrow-32+2x^2-16x^2+x^4=0\\ \Leftrightarrow-2\cdot\left(16-x^2\right)-x^2\cdot\left(16-x^2\right)=0\\ \Leftrightarrow\left(-2-x^2\right)\cdot\left(16-x^2\right)=0\\ \Leftrightarrow\left(-2-x^2\right)\cdot\left(4-x\right)\cdot\left(4+x\right)=0\\ \Rightarrow\left[{}\begin{matrix}-2-x^2=0\\4-x=0\\4+x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x^2=-2\left(ktm\right)\\x=4\\x=-4\end{matrix}\right.\)
(ktm) : không thỏa mãn
6)
\(x^4+2x^3-3x^2-8x-4=0\\ \Leftrightarrow x^4+x^3+x^3+x^2-4x^2-4x-4x-4=0\\ \Leftrightarrow\left(x^4+x^3\right)+\left(x^3+x^2\right)-\left(4x^2+4x\right)-\left(4x+4\right)=0\\ \Leftrightarrow x^2\cdot\left(x+1\right)-4x\cdot\left(x+1\right)-4\cdot\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\cdot\left(x^2-4x-4\right)=0\\ \Leftrightarrow\left(x+1\right)\cdot-\left(x^2+4x+4\right)=0\\ \Leftrightarrow\left(x+1\right)\cdot-\left(x+2\right)^2=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\-\left(x+2\right)^2=0\Leftrightarrow x+2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)
Bài 3.
1)
\(6x^2-5x+3=2x-3x\cdot\left(3-2x\right)\\ \Leftrightarrow6x^2-5x+3-2x+9x-6x^2=0\\ \Leftrightarrow2x+3=0\\ \Rightarrow x=-\frac{3}{2}\)
2)
\(\frac{2\cdot\left(x-4\right)}{4}-\frac{3+2x}{10}=x+\frac{1-x}{5}\\ \Leftrightarrow\frac{10\cdot\left(x-4\right)}{20}-\frac{2\cdot\left(3+2x\right)}{20}=\frac{20x}{20}+\frac{4\cdot\left(1-x\right)}{20}\\ \Leftrightarrow\frac{10x-40-6-4x}{20}=\frac{20x+4-4x}{20}\\ \Leftrightarrow\frac{6x-46}{20}=\frac{16x+4}{20}\\ \Leftrightarrow\frac{6x-46-16x-4}{20}=0\\ \Leftrightarrow\frac{-10x-50}{20}=0\\ \Rightarrow-10x-50=0\\ \Rightarrow x=\frac{50}{-10}=-5\)
3)
\(\frac{2x}{3}+\frac{3x-5}{4}=\frac{3\cdot\left(2x-1\right)}{2}-\frac{7}{6}\\ \Leftrightarrow\frac{8x}{12}+\frac{9x-15}{12}=\frac{36x-18}{12}-\frac{14}{12}\\ \Leftrightarrow\frac{8x+9x-15}{12}=\frac{36x-18-14}{12}\\ \Leftrightarrow\frac{17x-15}{12}=\frac{36x-32}{12}\\ \Leftrightarrow\frac{17x-15-36x+32}{12}=0\\ \Leftrightarrow\frac{17-19x}{12}=0\\ \Rightarrow17-19x=0\\ \Rightarrow x=\frac{17}{19}\)
4)
\(\frac{6x+5}{2}-\frac{10x+3}{4}=2x+\frac{2x+1}{2}\\ \Leftrightarrow\frac{12x+10}{4}-\frac{10x+3}{4}=\frac{8x}{4}+\frac{4x+2}{4}\\ \Leftrightarrow\frac{12x+10-10x-3}{4}=\frac{8x+4x+2}{4}\\ \Leftrightarrow\frac{2x+7}{4}=\frac{12x+2}{4}\\ \Leftrightarrow\frac{2x+7-12x-2}{4}=0\\ \Leftrightarrow\frac{5-10x}{4}=0\\ \Rightarrow5-10x=0\\ \Rightarrow x=\frac{5}{10}=\frac{1}{2}\)
5)
\(\left(x-4\right)\cdot\left(x+4\right)-2\cdot\left(3x-2\right)=\left(x-4\right)^2\\ \Leftrightarrow x^2-16-6x+4-x^2+8x-16=0\\ \Leftrightarrow2x-28=0\\ \Rightarrow x=\frac{28}{2}=14\)
6)
\(\left(x+1\right)^3-\left(x-1\right)^3=6\cdot\left(x^2+x+1\right)\\ \Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2-6x-6=0\\ \Leftrightarrow-6x-4=0\\ \Rightarrow x=\frac{4}{-6}=-\frac{2}{3}\)
