GIẢI CÁC PHƯƠNG TRÌNH
a) \(x-\frac{5x+2}{6}=\frac{7-3x}{4}\)
b) \(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)
c) \(\frac{x}{3}-\frac{2x+1}{2}=\frac{x}{6}-x\)
d) \(\frac{2+x}{3}=\frac{1}{2}x=\frac{1-2x}{4}+\frac{1}{4}\)
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HH
3 tháng 7 2020
a) 4 ( x + 5 )( x + 6 )( x + 10 )( x + 12 ) = 3x2
Do x = 0 không là nghiệm pt nên chia 2 vế pt cho \(x^2\ne0\), ta được :
\(\frac{4}{x^2}\left(x^2+60+17x\right)\left(x^2+60+16x\right)=3\)
\(\Leftrightarrow4\left(x+\frac{60}{x}+17\right)\left(x+\frac{60}{x}+16\right)=3\)
Đến đây ta đặt \(x+\frac{60}{x}+16=t\left(1\right)\)
Ta được :
\(4t\left(t+1\right)=3\Leftrightarrow4t^2+4t-3=0\Leftrightarrow\left(2t+3\right)\left(2t-1\right)=0\)
Từ đó ta lắp vào ( 1 ) tính được x
16 tháng 12 2022
a: \(\Leftrightarrow4\left(x^2+60+17x\right)\left(x^2+60+16x\right)=3x^2\)
\(\Leftrightarrow4\cdot\left[\left(x^2+60\right)^2+33x\left(x^2+60\right)+272x^2\right]=3x^2\)
=>4(x^2+60)^2+132x(x^2+60)+1085x^2=0
=>4(x^2+60)^2+62x(x^2+60)+70x(x^2+60)+1085x^2=0
=>2(x^2+60)(2x^2+120+31x)+35x(2x^2+120+31x)=0
=>(2x^2+120+35x)(2x^2+31x+120)=0
=>\(x\in\left\{\dfrac{-35\pm\sqrt{265}}{4};-\dfrac{15}{2};-8\right\}\)
b: Đặt x^2-3x=a
Phương trình sẽ là \(\dfrac{1}{a+3}+\dfrac{2}{a+4}=\dfrac{6}{a+5}\)
\(\Leftrightarrow\dfrac{a+4+2a+6}{\left(a+3\right)\left(a+4\right)}=\dfrac{6}{a+5}\)
=>(3a+10)(a+5)=6(a^2+7a+12)
=>6a^2+42a+72=3a^2+15a+10a+50
=>3a^2+17a+22=0
=>x=-2 hoặc x=-11/3
26 tháng 3 2020
a)
\(\frac{x}{3}-\frac{5x}{6}-\frac{15x}{12}=\frac{x}{4}-5\)
\(\Leftrightarrow\frac{4x-10x-15x}{12}=\frac{3x-60}{12}\)
\(\Leftrightarrow\frac{-10x-11}{12}=\frac{3x-60}{12}\)
\(\Leftrightarrow\frac{-10x-11-3x+60}{12}=0\)
\(\Leftrightarrow\frac{49-13x}{12}=0\)
\(\Rightarrow49-13x=0\)
\(\Rightarrow x=\frac{-49}{13}\)
26 tháng 3 2020
b)
\(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
\(\Leftrightarrow\frac{8x-3-6x+4}{4}=\frac{4x-2+x+3}{4}\)
\(\Leftrightarrow\frac{2x+1}{4}=\frac{5x+1}{4}\)
\(\Leftrightarrow\frac{2x+1-5x-1}{4}=0\)
\(\Leftrightarrow\frac{-3x}{4}=0\)
\(\Rightarrow-3x=0\)
\(\Rightarrow x=0\)
KN
18 tháng 8 2020
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. \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
\(\Leftrightarrow8x-3-2\left(3x-2\right)=2\left(2x-1\right)+x+3\)
\(\Leftrightarrow8x-3-6x+4=4x-2+x+3\)
\(\Leftrightarrow2x+1=5x+1\)
\(\Leftrightarrow2x=5x\)
\(\Leftrightarrow x=0\)
19 tháng 8 2020
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
24 tháng 1 2022
d: =>4x+6=15x-12
=>4x-15x=-12-6=-18
=>-11x=-18
hay x=18/11
e: =>\(45x+27=12+24x\)
=>21x=-15
hay x=-5/7
f: =>35x-5=96-6x
=>41x=101
hay x=101/41
g: =>3(x-3)=90-5(1-2x)
=>3x-9=90-5+10x
=>3x-9=10x+85
=>-7x=94
hay x=-94/7
a, \(x-\frac{5x+2}{6}=\frac{7-3x}{4}\)
\(\frac{12x}{12}-\frac{2\left(5x+2\right)}{12}=\frac{3\left(7-3x\right)}{12}\)
\(12x-10x-4=21-9x\)
\(11x=25\)
\(x=\frac{24}{11}\)
\(b,\frac{10x+3}{12}=1+\frac{6+8x}{9}\)
\(\frac{10x+3}{12}=\frac{15+8x}{9}\)
\(9\left(10x+3\right)=12\left(15+8x\right)\)
\(3\left(10x+3\right)=4\left(8x+15\right)\)
\(30x+9=32x+60\)
\(-2x=51\)
\(x=-\frac{51}{2}\)
\(c,\frac{x}{3}-\frac{2x+1}{2}=\frac{x}{6}-x\)
\(\frac{2x}{6}-\frac{3\left(2x+1\right)}{6}=\frac{x-6x}{6}\)
\(2x-6x-3=x-6x\)
\(x=3\)
P/s: Bn xem lại đề bài phần d nha!
=.= hk tốt!!
câu d sao bn đề đúng r ạ