Giải PT: \(\frac{x^2+2x+2}{x+1}+\frac{x^2+14x+56}{x+7}=\frac{x^2+6x+12}{x+3}+\frac{x^2+10x+30}{x+5}\)
Giai các pt sau:
\(\frac{1}{x^2+2x}+\frac{1}{x^2+6x+8}+\frac{1}{x^2+10x+24}+\frac{1}{x^2+14x+48}=\frac{4}{105}\)
\(ĐKXĐ:x\ne0;-2;-4;-6;-8\)\(\frac{1}{x\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+8\right)}=\frac{4}{105}\)
\(\Leftrightarrow\frac{2}{x\left(x+2\right)}+\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{2}{\left(x+4\right)\left(x+6\right)}+\frac{2}{\left(x+6\right)\left(x+8\right)}=\frac{8}{105}\)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+4}+...+\frac{1}{x+6}-\frac{1}{x+8}=\frac{8}{105}\)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+8}=\frac{8}{105}\)
Quy đồng làm nốt
bài 1 giải phương trình
a) (2x+3)\(^2\)-3(x-4)(x+4)=\(\left(x-2\right)^2\)+1
b)(3x-2) (9x\(^2\)+6x+4)-(3x-1) (9x\(^2\)+3x+1)=x-4
c)x (x-1) -(x-3) (x+4)=5x
d) (2x+1)(2x-1)=4x(x-7)-3x
bài 2 giải phương trình
a)\(\frac{x}{10}-\left(\frac{x}{30}+\frac{2x}{45}\right)=\frac{4}{5}\)
b)\(\frac{10x-5}{18}+\frac{x+3}{12}=\frac{7x+3}{6}+\frac{12-x}{9}\)
c)\(\frac{10x+3}{8}=\frac{7-8x}{12}\)
d)\(\frac{x+4}{5}-x-5=\frac{x+3}{3}-\frac{x-2}{2}\)
\(\frac{1}{x^2+2x}+\frac{1}{x^2+6x+18}+\frac{1}{x^2+10x+24}+\frac{1}{x^2+14x+48}=\frac{4}{150}\)
GIẢI PHƯƠNG TRÌNH......GIÚP MK NHA
Sai đề bạn ơi
Sai chỗ \(\frac{1}{x^2+6x+18}\)
Bạn sửa lại rồi mk giải cho..
giải phương trình:
\(\frac{x^2+4x+6}{x+2}-\frac{x^2+6x+12}{x+3}=\frac{x^2+8x+20}{x+4}-\frac{x^2+10x+30}{x+5}-\frac{4}{x^2+7x+12}\)
ko biết ok
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 phương trình:
\(\frac{1}{x^2+2x}+\frac{1}{x^2+6x+8}+\frac{1}{x^2+10x+24}+\frac{1}{x^2+14x+48}=\frac{4}{105}\)
PT<=> \(\frac{1}{x\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+8\right)}=\frac{4}{105}\)
<=> \(\frac{2}{x\left(x+2\right)}+\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{2}{\left(x+4\right)\left(x+6\right)}+\frac{2}{\left(x+6\right)\left(x+8\right)}=\frac{8}{105}\)
<=> \(\frac{1}{x}-\frac{1}{x+2}+\frac{1}{x+2}-...+\frac{1}{x+6}-\frac{1}{x+8}=\frac{8}{105}\)
<=> \(\frac{1}{x}-\frac{1}{x+8}=\frac{8}{105}\)
<=> \(\frac{8}{x\left(x+8\right)}=\frac{8}{105}\)
<=> x(x+8) = 105
<=> x = 7
mấy chế ai biết giải thì giải dùm mik mấy bài nè vs.
Giải phương trình:
1) \(\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}=\frac{1}{6}\)
2) \(\frac{1}{x^2-6x+8}+\frac{1}{x^2-10x+24}+\frac{1}{x^2-14x+48}=\frac{1}{9}\)
3) \(\frac{1}{x^2-3x+3}+\frac{2}{x^2-3x+4}=\frac{6}{x^2-3x+5}\)
4) \(\frac{6}{\left(x+1\right)\left(x+2\right)}+\frac{8}{\left(x-1\right)\left(x+4\right)}=1\)
5) \(4\left(x^3+\frac{1}{x^3}\right)=13\left(x+\frac{1}{x}\right)\)
6) \(\frac{4x}{4x^2-8x+7}+\frac{3x}{4x^2-10x+7}=1\)
7) \(\frac{x^2-3x+5}{x^2-4x+5}-\frac{x^2-5x+5}{x^2-6x+5}=\frac{-1}{4}\)
8) \(x\frac{8-x}{x-1}.\left(x-\frac{8-x}{x-1}\right)=15\)
1) Giải các pt:
a) 3(x - 1) - 2(x + 3)= -15
b) 3(x - 1) + 2= 3x - 1
c) 7(2 - 5x) - 5= 4(4 -6x)
2) Giải các pt phân thức: ( Tìm mẫu chung )
a) \(\frac{x}{30}+\frac{5x-1}{10}=\frac{x-8}{15}-\frac{2x+3}{6}\)
b) \(\frac{x+4}{5}-x+4=\frac{x}{3}-\frac{x-2}{2}\)
a, Ta có : \(3\left(x-1\right)-2\left(x+3\right)=-15\)
=> \(3x-3-2x-6=-15\)
=> \(3x-3-2x-6+15=0\)
=> \(x=-6\)
Vậy phương trình có nghiệm là x = -6 .
b, Ta có : \(3\left(x-1\right)+2=3x-1\)
=> \(3x-3+2=3x-1\)
=> \(3x-3+2-3x+1=0\)
=> \(0=0\)
Vậy phương trình có vô số nghiệm .
c, Ta có : \(7\left(2-5x\right)-5=4\left(4-6x\right)\)
=> \(14-35x-5=16-24x\)
=> \(14-35x-5-16+24x=0\)
=> \(-35x+24x=7\)
=> \(x=\frac{-7}{11}\)
Vậy phương trình có nghiệm là \(x=\frac{-7}{11}\) .
Bài 2 :
a, Ta có : \(\frac{x}{30}+\frac{5x-1}{10}=\frac{x-8}{15}-\frac{2x+3}{6}\)
=> \(\frac{x}{30}+\frac{3\left(5x-1\right)}{30}=\frac{2\left(x-8\right)}{30}-\frac{5\left(2x+3\right)}{30}\)
=> \(x+3\left(5x-1\right)=2\left(x-8\right)-5\left(2x+3\right)\)
=> \(x+15x-3=2x-16-10x-15\)
=> \(x+15x-3-2x+16+10x+15=0\)
=> \(24x+28=0\)
=> \(x=\frac{-28}{24}=\frac{-7}{6}\)
Vậy phương trình có nghiệm là \(x=\frac{-7}{6}\) .
b, Ta có : \(\frac{x+4}{5}-x+4=\frac{x}{3}-\frac{x-2}{2}\)
=> \(\frac{6\left(x+4\right)}{30}-\frac{30x}{30}+\frac{120}{30}=\frac{10x}{30}-\frac{15\left(x-2\right)}{30}\)
=> \(6\left(x+4\right)-30x+120=10x-15\left(x-2\right)\)
=> \(6x+24-30x+120=10x-15x+30\)
=> \(6x+24-30x+120-10x+15x-30=0\)
=> \(-19x+114=0\)
=> \(x=\frac{-114}{-19}=6\)
Vậy phương trình có nghiệm là x = 6 .
1) Giải các phương trình:
a) \(\frac{x-3}{5}-\frac{2x-1}{10}=\frac{x+1}{2}+\frac{1}{4}\)
b)\(\frac{x+3}{2}-\frac{2-1}{3}-1=\frac{x+5}{6}\)
c)\(\frac{x-1}{4}-\frac{5-2x}{9}=3x-\frac{2}{3}\)
d)\(\frac{2x-1}{4}+\frac{x-3}{3}=\frac{4x-2}{3}-\frac{6x+7}{12}\)
e)\(\frac{3x-2}{5}+\frac{x-1}{9}=\frac{14x-3}{15}-\frac{2x+1}{9}\)
\(\frac{x-3}{5}-\frac{2x-1}{10}=\frac{x+1}{2}+\frac{1}{4}\)
\(< =>\frac{\left(x-3\right).4}{20}-\frac{\left(2x-1\right).2}{20}=\frac{\left(x+1\right).10}{20}+\frac{5}{20}\)
\(< =>4x-12-4x+2=10x+10+5\)
\(< =>10x=-10-10-5=-25\)
\(< =>x=-\frac{25}{10}=-\frac{5}{2}\)
\(\frac{x+3}{2}-\frac{2x-1}{3}-1=\frac{x+5}{5}\)
\(< =>\frac{\left(x+3\right).15}{30}-\frac{\left(2x-1\right).10}{30}-\frac{30}{30}=\frac{\left(x+5\right).5}{30}\)\(< =>15x+45-20x+10-30=5x+25\)
\(< =>-5x+25=5x+25< =>10x=0< =>x=0\)
\(\frac{x-1}{4}-\frac{5-2x}{9}=3x-\frac{2}{3}\)
\(< =>\frac{\left(x-1\right).9}{36}-\frac{\left(5-2x\right).4}{36}=\frac{3x.36}{36}-\frac{2.12}{36}\)
\(< =>\left(x-1\right).9-\left(5-2x\right).4=108x-24\)
\(< =>9x-9-20+8x=108x-24\)
\(< =>108x-17x=-29+24\)
\(< =>91x=-5< =>x=-\frac{5}{91}\)