giải phương trình
a) \(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)
b) \(\frac{13}{2x^2+x-21}+\frac{1}{2x+7}=\frac{6}{x^2-9}\)
1, giải các phương trình sau
a, \(\frac{13}{2x^2+x-21}+\frac{1}{2x+7}=\frac{6}{x^2-9}\)
b, \(\frac{x-3}{x-5}+\frac{1}{x}=\frac{x+5}{x\left(x-5\right)}\)
c, \(\frac{1}{x+2}+\frac{1}{x^2-2x}=\frac{8}{x^3-4x}\)
d, \(\frac{2}{x^2-4}-\frac{1}{x^2-2x}=\frac{4+x}{x\left(x+2\right)}\)
a,\(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{x^2-9}\)\(\Leftrightarrow\frac{13\left(x+3\right)}{\left(x^2-9\right)\left(2x+7\right)}+\frac{x^2-9}{\left(x^2-9\right)\left(2x+7\right)}-\frac{6\left(2x+7\right)}{\left(x^2-9\right)\left(2x+7\right)}=0\)
\(\Leftrightarrow x+x^2-12=0\Leftrightarrow\orbr{\begin{cases}x=-4\\x=3\end{cases}}\)
b,\(\frac{x-3}{x-5}+\frac{1}{x}=\frac{x+5}{x\left(x-5\right)}\Leftrightarrow\frac{x\left(x-3\right)}{x\left(x-5\right)}+\frac{x-5}{x\left(x-5\right)}-\frac{x+5}{x\left(x-5\right)}=0\)
\(\Leftrightarrow x^2-3x-10=0\Rightarrow\orbr{\begin{cases}x=5\left(L\right)\\x=-2\end{cases}}\)
c,\(\frac{1}{x+2}+\frac{1}{x\left(x-2\right)}-\frac{8}{x\left(x^2-4\right)}=0\)\(\Leftrightarrow\frac{x\left(x-2\right)}{x\left(x^2-4\right)}+\frac{x+2}{x\left(x^2-4\right)}-\frac{8}{x\left(x^2-4\right)}=0\)
\(\Leftrightarrow x^2-x-6=0\Rightarrow\orbr{\begin{cases}x=3\\x=-2\left(L\right)\end{cases}}\)
d,\(\frac{2}{\left(x^2-4\right)}-\frac{1}{x\left(x-2\right)}-\frac{x+4}{x\left(x+2\right)}=0\)\(\Leftrightarrow\frac{2x}{x\left(x^2-4\right)}-\frac{x+2}{x\left(x^2-4\right)}-\frac{\left(x+4\right)\left(x-2\right)}{x\left(x^2-4\right)}=0\)
\(\Leftrightarrow-x^2-5x-10=0\)(vô nghiệm)
\(\)
Bài 1. Giải các phương trình sau
1) \(\frac{3x+2}{2}-\frac{3x+1}{6}=\frac{5}{3}-2x\)
2) \(\frac{x-3}{5}=6-\frac{1-2x}{3}\)
3) \(2\left(x+\frac{3}{5}\right)=5-\left(\frac{13}{5}+x\right)\)
4) \(\frac{2x+3}{3}=\frac{5-4x}{2}\)
5) \(\frac{5x+3}{12}=\frac{1+2x}{9}\)
6) \(x-\frac{x+1}{3}=\frac{2x+1}{5}\)
7) \(\frac{3\left(x-3\right)}{4}+\frac{4x-10,5}{10}=\frac{3\left(x+1\right)}{5}+6\)
8) \(\frac{2\left(3x+1\right)+1}{4}-5=\frac{2 \left(3x-1\right)}{5}-\frac{3x+2}{10}\)
9) \(\frac{x+1}{3}+\frac{3\left(2x+1\right)}{4}=\frac{2x+3\left(x+1\right)}{6}+\frac{7+12x}{12}\)
10) \(\frac{2x-1}{3}-\frac{5x+2}{7}=x+13\)
Giải phương trình:
1) \(\frac{7x-1}{6}+2x=\frac{16-x}{5}\)
2) \(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)
3)\(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
4)\(\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}=\frac{1-x}{2}-2\)
5)\(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)
6)\(\frac{6x+5}{2}-\left(2x+\frac{2x+1}{2}\right)=\frac{10x+3}{4}\)
7)\(\frac{2x-1}{5}-\frac{x-2}{3}-\frac{x+7}{15}=0\)
8)\(\frac{x+4}{5}-x+4=\frac{x}{3}-\frac{x-2}{2}\)
giải giúp mik với ạ
1. \(\frac{7x-1}{6}+2x=\frac{16-x}{5}\)
\(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)
\(\Leftrightarrow35x-5+60x=96-6x\)
\(\Leftrightarrow95x-5=96-6x\)
\(\Leftrightarrow95x+6x=96+5\)
\(\Leftrightarrow101x=101\)
\(\Leftrightarrow x=1\)
2. \(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)
\(\Leftrightarrow3\left(10x+3\right)=36+4\left(6+8x\right)\)
\(\Leftrightarrow30x+9=36+24+32x\)
\(\Leftrightarrow30x+9=32x+60\)
\(\Leftrightarrow30x-32x=60-9\)
\(\Leftrightarrow-2x=51\)
\(\Leftrightarrow x=-\frac{51}{2}\)
3. \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
\(\Leftrightarrow8x-3-2\left(3x-2\right)=2\left(2x-1\right)+x+3\)
\(\Leftrightarrow8x-3-6x+4=4x-2+x+3\)
\(\Leftrightarrow2x+1=5x+1\)
\(\Leftrightarrow2x=5x\)
\(\Leftrightarrow x=0\)
4) \(\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}=\frac{1-x}{2}-2\)
=> \(\frac{9-3x}{8}+\frac{10-2x}{3}=\frac{1-x}{2}-\frac{2}{1}\)
=> \(\frac{3\left(9-3x\right)}{24}+\frac{8\left(10-2x\right)}{24}=\frac{12\left(1-x\right)}{24}-\frac{48}{24}\)
=> \(\frac{27-9x}{24}+\frac{80-16x}{24}=\frac{12-12x}{24}-\frac{48}{24}\)
=> \(\frac{27-9x+80-16x}{24}=\frac{12-12x-48}{24}\)
=> 27 - 9x + 80 - 16x = 12 - 12x - 48
=> 27 - 9x + 80 - 16x - 12 + 12x + 48 = 0
=> (27 + 80 - 12 + 48) + (-9x - 16x + 12x) = 0
=> 143 - 13x = 0
=> 13x = 143
=> x = 11
5) \(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)
=> \(\frac{2x-6}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)
=> \(\frac{3\left(2x-6\right)}{21}+\frac{7\left(x-5\right)}{21}-\frac{13x+4}{21}=0\)
=> \(\frac{6x-18}{21}+\frac{7x-35}{21}-\frac{13x+4}{21}=0\)
=> \(\frac{6x-18+7x-35-13x-4}{21}=0\)
=> 6x - 18 + 7x - 35 - 13x - 4 = 0
=> (6x + 7x - 13x) + (-18 - 35 - 4) = 0
=> -57 = 0(vô nghiệm)
6) \(\frac{6x+5}{2}-\left(2x+\frac{2x+1}{2}\right)=\frac{10x+3}{4}\)
=> \(\frac{6x+5}{2}-\frac{10x+3}{4}=2x+\frac{2x+1}{2}\)
=> \(\frac{2\left(6x+5\right)}{4}-\frac{10x+3}{4}=\frac{8x}{4}+\frac{2\left(2x+1\right)}{4}\)
=> \(\frac{12x+10}{4}-\frac{10x+3}{4}=\frac{8x}{4}+\frac{4x+2}{4}\)
=> \(\frac{12x+10-\left(10x+3\right)}{4}=\frac{8x+4x+2}{4}\)
=> \(\frac{12x+10-10x-3}{4}=\frac{12x+2}{4}\)
=> \(12x+10-10x-3=12x+2\)
=> \(2x+10-3=12x+2\)
=> 2x + 10 - 3 - 12x - 2 = 0
=> (2x - 12x) + (10 - 3 - 2) = 0
=> -10x + 5 = 0
=> -10x = -5
=> x = 1/2
7) \(\frac{2x-1}{5}-\frac{x-2}{3}-\frac{x+7}{15}=0\)
=> \(\frac{3\left(2x-1\right)}{15}-\frac{5\left(x-2\right)}{15}-\frac{x+7}{15}=0\)
=> \(\frac{6x-3}{15}-\frac{5x-10}{15}-\frac{x+7}{15}=0\)
=> \(\frac{6x-3-\left(5x-10\right)-\left(x+7\right)}{15}=0\)
=> 6x - 3 - 5x + 10 - x - 7 = 0
=> (6x - 5x - x) + (-3 + 10 - 7) = 0
=> 0x + 0 = 0
=> 0x = 0
=> x tùy ý
Bài 8 tự làm nhé
Giải các phương trình sau:
a) \(\frac{x+1}{x-2}+\frac{x-1}{x+2}=\frac{2.\left(x^2+2\right)}{x^2-4}\)
b) \(\frac{5-x}{4x^2-8x}-\frac{1}{8x-16}=\frac{x-1}{2x^2-4x}-\frac{7}{8x}\)
c) \(\frac{13}{2x^2+x-21}+\frac{1}{2x+7}=\frac{6}{x^2-9}\)
d)\(\frac{x-3}{x-2}+\frac{x-2}{x-4}=\frac{11}{3}\)
e) \(\frac{2.x}{x-1}=\frac{5.\left(x-1\right)}{x+1}\)
f) \(\frac{1}{x^2-2.x+2}+\frac{2}{x^2-2.x+3}=\frac{6}{x^2-2.x+4}\)
Ai giúp mình được bài nào thì giúp nha. Cám ơn
a.2x#+_2 . quy đồng khử mẫu tchung : (x+2)(x+1)+(x-1)(x-2)--->2x^2 + 4=2(x^2+2). --> s={x thuộc R/ X#+_2}
a) ĐKXĐ \(\hept{\begin{cases}x\ne-2\\x\ne2\end{cases}}\)
\(\Rightarrow\left(x+1\right)\left(x+2\right)+\left(x-1\right)\left(x-2\right)-2x\left(x^2+2\right)=0\)
\(\Leftrightarrow x^2+3x+2+x^2-3x+2-2x^2-4=0\)
\(\Leftrightarrow0x=0\)(vô số nghiệm)
nghiệm x thỏa mãn phương trình S \(\in\)R với \(\hept{\begin{cases}x\ne-2\\x\ne2\end{cases}}\)
b) ĐKXĐ \(\hept{\begin{cases}x\ne0\\x\ne2\end{cases}}\)
\(\Rightarrow\frac{5-x}{4x\left(x-2\right)}-\frac{1}{8\left(x-2\right)}=\frac{1}{2x\left(x-2\right)}-\frac{7}{8x}\)
\(\Rightarrow2\left(5-x\right)-x-4\left(x-1\right)+7\left(x-2\right)=0\)
\(\Leftrightarrow10-2x-x-4x+4+7x-14=0\)
\(\Leftrightarrow0x=0\)(phương trìh vô số nghiệm)
nghiệm x thỏa mãn phương trình S \(\in\)R với \(\hept{\begin{cases}x\ne0\\x\ne2\end{cases}}\)
1) giải phương trình:
a) \(\left(2x+3\right)\left(\frac{3x+8}{2-7x}+1\right)=\left(x+5\right)\left(\frac{3x+8}{2-7x}+1\right)\)
b) \(\frac{7x+10}{x+1}\left(x^2-x-2\right)-\frac{7x+10}{x+1}\left(2x^2-3x-5\right)=0\)
c) \(\frac{2x+5}{x+3}+1=\frac{4}{x^2+2x-3}-\frac{3x-1}{1-x}\)
d) \(\frac{13}{2x^2+x-21}+\frac{1}{2x+7}+\frac{6}{9-x^2}=0\)
e) \(\frac{x-49}{50}+\frac{x-50}{49}=\frac{49}{x-50}+\frac{50}{x-49}\)
f) \(\frac{1+\frac{x}{x+3}}{1-\frac{x}{x+3}}=3\)
giải phương trình
a \(\frac{1}{x^2-2x+2}+\frac{2}{x^2-2x+3}=\frac{6}{x^2-2x+4}\)
b \(\frac{x^2+2x+1}{x^2+2x+2}+\frac{x^2+2x+2}{x^2+2x+3}=\frac{7}{6}\)
c \(x^2+\frac{1}{x^2}-\frac{9}{2}\left(x+\frac{1}{x}\right)+7=0\)
Hướng dẫn:
a) Đặt : \(x^2-2x+1=t\)Ta có:
\(\frac{1}{t+1}+\frac{2}{t+2}=\frac{6}{t+3}\)
b) Đặt : \(x^2+2x+1=t\)
Ta có pt: \(\frac{t}{t+1}+\frac{t+1}{t+2}=\frac{7}{6}\)
c)ĐK: x khác 0
Đặt: \(x+\frac{1}{x}=t\)
KHi đó: \(x^2+\frac{1}{x^2}=t^2-2\)
Ta có pt: \(t^2-2-\frac{9}{2}t+7=0\)
a) Đặt \(x^2-2x+3=v\)
Phương trình trở thành \(\frac{1}{v-1}+\frac{2}{v}=\frac{6}{v+1}\)
\(\Rightarrow\frac{v\left(v+1\right)+2\left(v+1\right)\left(v-1\right)}{v\left(v+1\right)\left(v-1\right)}=\frac{6v\left(v-1\right)}{v\left(v+1\right)\left(v-1\right)}\)
\(\Rightarrow v\left(v+1\right)+2\left(v+1\right)\left(v-1\right)=6v\left(v-1\right)\)
\(\Rightarrow v^2+v+2v^2-2=6v^2-6v\)
\(\Rightarrow3v^2-7v+2=0\)
Ta có \(\Delta=7^2-4.3.2=25,\sqrt{\Delta}=5\)
\(\Rightarrow\orbr{\begin{cases}v=\frac{7+5}{6}=2\\v=\frac{7-5}{6}=\frac{1}{3}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x^2-2x+3=2\\x^2-2x+3=\frac{1}{3}\end{cases}}\)
+) \(x^2-2x+1=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)
+)\(x^2-2x+3=\frac{1}{3}\)
\(\Rightarrow x^2-2x+\frac{8}{3}=0\)
Ta có \(\Delta=2^2-4.\frac{8}{3}=\frac{-20}{3}< 0\)
Vậy phương trình có 1 nghiệm là x = 1
c) Đặt \(\left(x+\frac{1}{x}\right)=a\) Khi đó pt có dạng :
\(a^2-\frac{9}{2}a+7-2=0\)
\(\Leftrightarrow2a^2-9a+10=0\)
\(\Leftrightarrow2a^2-4a-5a+10=0\)
\(\Leftrightarrow\left(a-2\right)\left(2a-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=2\\a=\frac{5}{2}\end{cases}}\)
+) Với \(a=\frac{5}{2}\Rightarrow x+\frac{1}{x}=\frac{5}{2}\)
\(\Rightarrow x^2+1=\frac{5x}{2}\)
\(\Rightarrow2x^2+2-5x=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{1}{2}\end{cases}}\) ( thỏa mãn)
+) Với \(a=2\Rightarrow x+\frac{1}{x}=2\)
\(\Rightarrow x^2-2x+1=0\)
\(\Leftrightarrow\left(x-1\right)^2=0\)
\(\Leftrightarrow x=1\) ( thỏa mãn )
Vậy pt đã cho có tập nghiệm \(S=\left\{1,\frac{1}{2},2\right\}\)
Tìm điều kiện xác định rồi giải các phương trình sau:
a) \(\frac{x-2}{2+x}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{x^2-4}\)
b) \(\frac{3}{4\left(x-5\right)}+\frac{15}{50-2x^2}=\frac{-7}{6\left(x+5\right)}\)
c) \(\frac{8x^2}{3\left(1-4x^2\right)}=\frac{2x}{6x-3}-\frac{1+8x}{4+8x}\)
d) \(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{x^2-9}\)
Help me!
a) ĐKXĐ: x khác +2
\(\frac{x-2}{2+x}-\frac{3}{x-2}-\frac{2\left(x-11\right)}{x^2-4}\)
<=> \(\frac{x-2}{2+x}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}\)
<=> (x - 2)^2 - 3(2 + x) = 2(x - 11)
<=> x^2 - 4x + 4 - 6 - 3x = 2x - 22
<=> x^2 - 7x - 2 = 2x - 22
<=> x^2 - 7x - 2 - 2x + 22 = 0
<=> x^2 - 9x + 20 = 0
<=> (x - 4)(x - 5) = 0
<=> x - 4 = 0 hoặc x - 5 = 0
<=> x = 4 hoặc x = 5
làm nốt đi
Dạng 1: Phương trình bậc nhất
Bài 1: Giải các phương trình sau :
a) 0,5x (2x - 9) = 1,5x (x - 5)
b) 28 (x - 1) - 9 (x - 2) = 14x
c) 8 (3x - 2) - 14x = 2 (4 - 7x) + 18x
d) 2 (x - 5) - 6 (1 - 2x) = 3x + 2
e) \(\frac{x+7}{2}-\frac{x-3}{5}=\frac{x}{6}\)
f) \(\frac{2x-3}{3}-\frac{5x+2}{12}=\frac{x-3}{4}+1\)
g) \(\frac{x+6}{2}+\frac{2\left(x+17\right)}{2}+\frac{5\left(x-10\right)}{6}=2x+6\)
h) \(\frac{3x+2}{5}-\frac{4x-3}{7}=4+\frac{x-2}{35}\)
i) \(\frac{x-1}{2}+\frac{x+3}{3}=\frac{5x+3}{6}\)
j) \(\frac{x-3}{5}-1=\frac{4x+1}{4}\)
Dạng 2: Phương trình tích
Bài 2: Giải phương trình sau :
a) (x + 1) (5x + 3) = (3x - 8) (x - 1)
b) (x - 1) (2x - 1) = x(1 - x)
c) (2x - 3) (4 - x) (x - 3) = 0
d) (x + 1)2 - 4x2 = 0
e) (2x + 5)2 = (x + 3)2
f) (2x - 7) (x + 3) = x2 - 9
g) (3x + 4) (x - 4) = (x - 4)2
h) x2 - 6x + 8 = 0
i) x2 + 3x + 2 = 0
j) 2x2 - 5x + 3 = 0
k) x (2x - 7) - 4x + 14 = 9
l) (x - 2)2 - x + 2 = 0
Dạng 3: Phương trình chứa ẩn ở mẫu
Bài 3: Giải phương trình sau :
\(\frac{90}{x}-\frac{36}{x-6}=2\) | \(\frac{3}{x+2}-\frac{2}{x-3}=\frac{8}{\left(x-3\right)\left(x+2\right)}\) |
\(\frac{1}{x}+\frac{1}{x+10}=\frac{1}{12}\) | \(\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\) |
\(\frac{x+3}{x-3}-\frac{1}{x}=\frac{3}{x\left(x-3\right)}\) | \(\frac{3}{4\left(x-5\right)}+\frac{15}{50-2x^2}=\frac{-7}{6\left(x+5\right)}\) |
\(\frac{3}{x+2}-\frac{2}{x-2}+\frac{8}{x^2-4}=0\) | \(\frac{x}{x+1}-\frac{2x-3}{1-x}=\frac{3x^2+5}{x^2-1}\) |
1 . \(\frac{x-1}{12}-\frac{2x-12}{14}=\frac{3x-14}{25}-\frac{4x-25}{27}\)
2 . \(\frac{3}{x-1}+\frac{4}{x-2}=\frac{5}{x-3}+\frac{6}{x-4}\)
3 . \(\frac{2x-1}{2}+\frac{2x+1}{3}=\frac{2x+7}{6}+\frac{2x+9}{7}\)
4 . \(\frac{x^2+x+4}{2}+\frac{x^2+x+7}{3}=\frac{x^2+x+13}{5}+\frac{x^2+x+16}{6}\)