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\(2x^4+3x^3+8x^2+6x+5=0\)
\(\Leftrightarrow2x^4+2x^3+2x^2+x^3+x^2+x+5x^2+5x+5=0\)
\(\Leftrightarrow2x^2\left(x^2+x+1\right)+x\left(x^2+x+1\right)+5\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\left(x^2+x+1\right)\left(2x^2+x+5\right)=0\)
Mà \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\)
\(2x^2+x+5=2\left[\left(x+\frac{1}{4}\right)^2+\frac{39}{16}\right]>0\forall x\)
Vậy tập nghiệm của pt là \(S=\varnothing\)
b, \(\frac{x-342}{15}+\frac{x-323}{17}+\frac{x-300}{19}+\frac{x-273}{21}=10\)
\(\Leftrightarrow\left(\frac{x-342}{15}-1\right)+\left(\frac{x-323}{17}-2\right)+\left(\frac{x-300}{19}-3\right)+\left(\frac{x-273}{21}-4\right)=0\)
\(\Leftrightarrow\frac{x-357}{15}+\frac{x-357}{17}+\frac{x-357}{19}+\frac{x-357}{21}=0\)
\(\Leftrightarrow\left(x-357\right)\left(\frac{1}{15}+\frac{1}{17}+\frac{1}{19}+\frac{1}{21}\right)=0\)
\(\Leftrightarrow x-357=0\Leftrightarrow x=357\)
Vậy tập nghiệm của pt: \(S=\left\{357\right\}\)
Câu a)
\(2x^4+3x^3+8x^2+6x+5=0\)
\(\Leftrightarrow (2x^4+2x^3+2x^2)+(x^3+x^2+x)+5x^2+5x+5=0\)
\(\Leftrightarrow 2x^2(x^2+x+1)+x(x^2+x+1)+5(x^2+x+1)=0\)
\(\Leftrightarrow (x^2+x+1)(2x^2+x+5)=0\)
\(\Rightarrow \left[\begin{matrix} x^2+x+1=0\\ 2x^2+x+5=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} (x+\frac{1}{2})^2+\frac{3}{4}=0\\ 2(x+\frac{1}{4})^2+\frac{39}{8}=0\end{matrix}\right.\) (vô lý)
Vậy pt vô nghiệm.
Cách khác:
PT \(\Leftrightarrow 4x^4+6x^3+16x^2+12x+10=0\)
\(\Leftrightarrow 3x^4+(x^4+6x^3+9x^2)+7x^2+12x+10=0\)
\(\Leftrightarrow 3x^4+(x^2+3x)^2+(4x^2+12x+9)+3x^2+1=0\)
\(\Leftrightarrow 3x^4+(x^2+3x)^2+(2x+3)^2+3x^2=-1\)
(vô lý vì vế phải âm còn vế trái không âm)
Vậy pt vô nghiệm.
Câu b:
\(\frac{x-342}{15}+\frac{x-323}{17}+\frac{x-300}{19}+\frac{x-273}{21}=10\)
\(\Leftrightarrow \frac{x-342}{15}+\frac{x-323}{17}+\frac{x-300}{19}+\frac{x-273}{21}-10=0\)
\(\Leftrightarrow \frac{x-342}{15}-1+\frac{x-323}{17}-2+\frac{x-300}{19}-3+\frac{x-273}{21}-4=0\)
\(\Leftrightarrow \frac{x-357}{15}+\frac{x-357}{17}+\frac{x-357}{19}+\frac{x-357}{21}=0\)
\(\Leftrightarrow (x-357)\left(\frac{1}{15}+\frac{1}{17}+\frac{1}{19}+\frac{1}{21}\right)=0\)
Dễ thấy \(\frac{1}{15}+\frac{1}{17}+\frac{1}{19}+\frac{1}{21}\neq 0\), do đó $x-357=0$ hay $x=357$ là nghiệm duy nhất của pt.
\(a,\left(x-2\right)\left(x-3\right)-3\left(4x-2\right)=\left(x-4\right)^2\\ \Leftrightarrow x^2-5x+6-12x+6=x^2-8x+16\\ \Leftrightarrow-9x-4=0\\ \Leftrightarrow x=-\dfrac{4}{9}\)
\(b,\dfrac{2x^2+1}{8}-\dfrac{7x-2}{12}=\dfrac{x^2-1}{4}-\dfrac{x-3}{6}\\ \Leftrightarrow6x^2+3-14x+4=6x^2-6-4x+12\\ \Leftrightarrow10x=1\\ \Leftrightarrow x=\dfrac{1}{10}\)
\(c,x-\dfrac{2x-2}{5}+\dfrac{x+8}{6}=7+\dfrac{x-1}{3}\\ \Leftrightarrow30x-12x+12+5x+40=210+10x-10\\ \Leftrightarrow13x=148\\ \Leftrightarrow x=\dfrac{148}{13}\)
\(d,\left(2x+5\right)^2=\left(x+2\right)^2\\ \Leftrightarrow\left(2x+5\right)^2-\left(x+2\right)^2=0\\ \Leftrightarrow\left(2x+5-x-2\right)\left(2x+5+x+2\right)=0\\ \Leftrightarrow\left(x+3\right)\left(3x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{7}{3}\end{matrix}\right.\)
\(e,x^2-5x+6=0\\ \Leftrightarrow\left(x-2\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
\(g,2x^3+6x^2=x^2+3x\\ \Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow x\left(2x-1\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=-3\end{matrix}\right.\)
\(h,\left(x+\dfrac{1}{x}\right)^2+2\left(x+\dfrac{1}{x}\right)-8=0\left(x\ne0\right)\)
Đặt \(x+\dfrac{1}{x}=t\), pt trở thành:
\(t^2+2t-8=0\\ \Leftrightarrow\left(t-2\right)\left(t+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}t=2\\t=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=2\\x+\dfrac{1}{x}=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1-2x=0\\x^2+1+4x=0\left(1\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\Delta\left(1\right)=16-4=12>0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\\left[{}\begin{matrix}x=-2+\sqrt{3}\\x=-2-\sqrt{3}\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2+\sqrt{3}\\x=-2-\sqrt{3}\end{matrix}\right.\)
Tick plzz
a) Ta có: \(\dfrac{x-3}{5}=6-\dfrac{1-2x}{3}\)
\(\Leftrightarrow\dfrac{3\left(x-3\right)}{15}=\dfrac{90}{15}-\dfrac{5\left(1-2x\right)}{15}\)
\(\Leftrightarrow3x-9=90-5+10x\)
\(\Leftrightarrow3x-9=10x+85\)
\(\Leftrightarrow3x-10x=85+9\)
\(\Leftrightarrow-7x=94\)
hay \(x=-\dfrac{94}{7}\)
Vậy: \(S=\left\{-\dfrac{94}{7}\right\}\)
b) Ta có: \(\dfrac{3x-2}{6}-5=\dfrac{3-2\left(x+7\right)}{4}\)
\(\Leftrightarrow\dfrac{2\left(3x-2\right)}{12}-\dfrac{60}{12}=\dfrac{3\left(3-2x-14\right)}{12}\)
\(\Leftrightarrow6x-4-60=9-6x-42\)
\(\Leftrightarrow6x-64=-6x-33\)
\(\Leftrightarrow6x+6x=-33+64\)
\(\Leftrightarrow12x=31\)
hay \(x=\dfrac{31}{12}\)
Vậy: \(S=\left\{\dfrac{31}{12}\right\}\)
c) Ta có: \(3\left(x-1\right)+3=5x\)
\(\Leftrightarrow3x-3+3=5x\)
\(\Leftrightarrow3x-5x=0\)
\(\Leftrightarrow-2x=0\)
hay x=0
Vậy: S={0}
d) Ta có: \(\dfrac{x+1}{100}+\dfrac{x+2}{99}=\dfrac{x+3}{98}+\dfrac{x+4}{97}\)
\(\Leftrightarrow\dfrac{x+1}{100}+1+\dfrac{x+2}{99}+1=\dfrac{x+3}{98}+1+\dfrac{x+4}{97}+1\)
\(\Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}=\dfrac{x+101}{98}+\dfrac{x+101}{97}\)
\(\Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}-\dfrac{x+101}{98}-\dfrac{x+101}{97}=0\)
\(\Leftrightarrow\left(x+101\right)\left(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\right)=0\)
mà \(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\ne0\)
nên x+101=0
hay x=-101
Vậy: S={-101}
a) \(\dfrac{x-3}{5}=6-\dfrac{1-2x}{3}\\ \Leftrightarrow\dfrac{3\left(x-3\right)}{15}=\dfrac{90-5\left(1-2x\right)}{15}\\ \Leftrightarrow3x-9=90-5+10x\\ \Leftrightarrow3x-10x=90-5+9\\ \Leftrightarrow-7x=94\\ \Leftrightarrow x=\dfrac{-94}{7}\)
Vậy \(x=\dfrac{-94}{7}\) là nghiệm của pt
b) \(\dfrac{3x-2}{6}-5=\dfrac{3-2\left(x+7\right)}{4}\\ \Leftrightarrow\dfrac{2\left(3x-2\right)-60}{12}=\dfrac{9-6\left(x+7\right)}{12}\\ \Leftrightarrow6x-4-60=9-6x-42\\ \Leftrightarrow6x+6x=9-42+4+60\\ \Leftrightarrow12x=31\\ \Leftrightarrow x=\dfrac{31}{12}\)
Vậy \(x=\dfrac{31}{12}\) là nghiệm của pt
c) \(3\left(x-1\right)+3=5x\\ \Leftrightarrow3x+3+3=5x\\ \Leftrightarrow5x-3x=3+3\\ \Leftrightarrow2x=6\\ \Leftrightarrow x=3\)
Vậy x = 3 là nghiệm của pt
d) \(\dfrac{x+1}{100}+\dfrac{x+2}{99}=\dfrac{x+3}{98}+\dfrac{x+4}{97}\\ \Leftrightarrow\left(\dfrac{x+1}{100}+1\right)+\left(\dfrac{x+2}{99}+1\right)=\left(\dfrac{x+3}{98}+1\right)+\left(\dfrac{x+4}{97}+1\right)\\ \Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}-\dfrac{x+101}{98}-\dfrac{x+101}{97}=0\\ \Leftrightarrow\left(x+101\right)\left(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\right)=0\\ \Leftrightarrow x+101=0\\ \Leftrightarrow x=-101\)
Vậy x = -101 là nghiệm của pt
e) \(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}=-4\\ \Leftrightarrow\left(\dfrac{59-x}{41}+1\right)+\left(\dfrac{57-x}{43}+1\right)+\left(\dfrac{53-x}{45}+1\right)+\left(\dfrac{53-x}{47}+1\right)=0\\ \Leftrightarrow\dfrac{100-x}{41}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}=0\\ \Leftrightarrow\left(100-x\right)\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}\right)=0\\ \Leftrightarrow100-x=0\\ \Leftrightarrow x=100\)
Vậy x = 100 là nghiệm của pt
f) \(\dfrac{x-90}{10}+\dfrac{x-76}{12}+\dfrac{x-58}{14}+\dfrac{x-36}{16}+\dfrac{x-15}{17}=15\\ \Leftrightarrow\left(\dfrac{x-90}{10}-1\right)+\left(\dfrac{x-76}{12}-2\right)+\left(\dfrac{x-58}{14}-3\right)+\left(\dfrac{x-36}{16}-4\right)+\left(\dfrac{x-15}{17}-5\right)=0\\ \Leftrightarrow\dfrac{x-100}{10}+\dfrac{x-100}{12}+\dfrac{x-100}{14}+\dfrac{x-100}{16}+\dfrac{x-100}{17}=0\\ \Leftrightarrow\left(x-100\right)\left(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+\dfrac{1}{17}\right)=0\\ \Leftrightarrow x-100=0\\ \Leftrightarrow x=100\)
Vậy x = 100 là nghiệm của pt
giải pt sau
g) 11+8x-3=5x-3+x
\(\Leftrightarrow\) 8x + 8 = 6x - 3
<=> 8x-6x = -3 - 8
<=> 2x = -11
=> x=-\(\dfrac{11}{2}\)
Vậy tập nghiệm của PT là : S={\(-\dfrac{11}{2}\)}
h)4-2x+15=9x+4-2x
<=> 19 - 2x = 7x + 4
<=> -2x - 7x = 4 - 19
<=> -9x = -15
=> x=\(\dfrac{15}{9}=\dfrac{5}{3}\)
Vậy tập nghiệm của pt là : S={\(\dfrac{5}{3}\)}
g)\(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=\dfrac{5}{3}+2x\)
<=> \(\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{5.2+6.2x}{6}\)
<=> 9x + 6 - 3x + 1 = 10 + 12x
<=> 6x + 7 = 10 + 12x
<=> 6x -12x = 10-7
<=> -6x = 3
=> x= \(-\dfrac{1}{2}\)
Vậy tập nghiệm của PT là : S={\(-\dfrac{1}{2}\)}
\(h,\dfrac{x+4}{5}-x+4=\dfrac{4x+2}{5}-5\)
<=> \(\dfrac{x+4-5\left(x+4\right)}{5}=\dfrac{4x+2-5.5}{5}\)
<=> x + 4 - 5x - 20 = 4x + 2 - 25
<=> x - 5x - 4x = 2-25-4+20
<=> -8x = -7
=> x= \(\dfrac{7}{8}\)
Vậy tập nghiệm của PT là S={\(\dfrac{7}{8}\)}
\(i,\dfrac{4x+3}{5}-\dfrac{6x-2}{7}=\dfrac{5x+4}{3}+3\)
<=> \(\dfrac{21\left(4x+3\right)}{105}\)-\(\dfrac{15\left(6x-2\right)}{105}\)=\(\dfrac{35\left(5x+4\right)+3.105}{105}\)
<=> 84x + 63 - 90x + 30 = 175x + 140 + 315
<=> 84x - 90x - 175x = 140 + 315 - 63 - 30
<=> -181x = 362
=> x = -2
Vậy tập nghiệm của PT là : S={-2}
K) \(\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)
<=> \(\dfrac{5\left(5x+2\right)}{30}-\dfrac{10\left(8x-1\right)}{30}=\dfrac{6\left(4x+2\right)-150}{30}\)
<=> 25x + 10 - 80x - 10 = 24x + 12 - 150
<=> -55x = 24x - 138
<=> -55x - 24x = -138
=> -79x = -138
=> x=\(\dfrac{138}{79}\)
Vậy tập nghiệm của PT là S={\(\dfrac{138}{79}\)}
m) \(\dfrac{2x-1}{5}-\dfrac{x-2}{3}=\dfrac{x+7}{15}\)
<=> \(\dfrac{3\left(2x-1\right)-5\left(x-2\right)}{15}=\dfrac{x+7}{15}\)
<=> 6x - 3 - 5x + 10 = x+7
<=> x + 7 = x+7
<=> 0x = 0
=> PT vô nghiệm
Vậy S=\(\varnothing\)
n)\(\dfrac{1}{4}\left(x+3\right)=3-\dfrac{1}{2}\left(x+1\right)-\dfrac{1}{3}\left(x+2\right)\)
<=> \(\dfrac{1}{4}x+\dfrac{3}{4}=3-\dfrac{1}{2}x-\dfrac{1}{2}-\dfrac{1}{3}x-\dfrac{2}{3}\)
<=> \(\dfrac{1}{4}x+\dfrac{1}{2}x+\dfrac{1}{3}x=3-\dfrac{1}{2}-\dfrac{2}{3}-\dfrac{3}{4}\)
<=> \(\dfrac{13}{12}x=\dfrac{13}{12}\)
=> x= 1
Vậy S={1}
p) \(\dfrac{x}{3}-\dfrac{2x+1}{6}=\dfrac{x}{6}-6\)
<=> \(\dfrac{2x-2x+1}{6}=\dfrac{x-36}{6}\)
<=> 2x -2x + 1= x-36
<=> 2x-2x-x = -37
=> x = 37
Vậy S={37}
q) \(\dfrac{2+x}{5}-0,5x=\dfrac{1-2x}{4}+0,25\)
<=> \(\dfrac{4\left(2+x\right)-20.0,5x}{20}=\dfrac{5\left(1-2x\right)+20.0,25}{20}\)
<=> 8 + 4x - 10x = 5 - 10x + 5
<=> 4x-10x + 10x = 5+5-8
<=> 4x = 2
=> x= \(\dfrac{1}{2}\)
Vậy S={\(\dfrac{1}{2}\)}
g) \(11+8x-3=5x-3+x\)
\(\Leftrightarrow8+8x=6x-3\)
\(\Leftrightarrow8x-6x=-3-8\)
\(\Leftrightarrow2x=-11\)
\(\Leftrightarrow x=-\dfrac{11}{2}\)
h, \(4-2x+15=9x+4-2x\)
\(\Leftrightarrow-2x-9x+2x=4-4-15\)
\(\Leftrightarrow-9x=-15\)
\(\Leftrightarrow x=\dfrac{-15}{-9}=\dfrac{5}{3}\)
a: =>5-x+6=12-8x
=>-x+11=12-8x
=>7x=1
hay x=1/7
b: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
\(\Leftrightarrow9x+6-3x-1=12x+10\)
=>12x+10=6x+5
=>6x=-5
hay x=-5/6
d: =>(x-2)(x-3)=0
=>x=2 hoặc x=3
bạn tách một câu vài câu hỏi chứ đừng gộp như thế này ko ai trả lời đâu
a: =>\(4x-5=2x-2+x=3x-2\)
=>x=3
b: \(\Leftrightarrow7x-35=3x+6\)
=>4x=41
=>x=41/4
c: =>(2x+5)(x+5)-2x^2=0
=>2x^2+10x+5x+25-2x^2=0
=>15x=-25
=>x=-5/3
e: \(\Leftrightarrow\dfrac{11}{x}=\dfrac{9x-36+2x+2}{\left(x+1\right)\left(x-4\right)}\)
=>11(x^2-3x-4)=x(11x-34)
=>11x^2-33x-44=11x^2-34x
=>x=44
\(1,\dfrac{4x-4}{3}=\dfrac{7-x}{5}\\ \Leftrightarrow5\left(4x-4\right)=3\left(7-x\right)\\ \Leftrightarrow20x-20=21-3x\\ \Leftrightarrow17x=41\Leftrightarrow x=\dfrac{41}{17}\)
\(2,\dfrac{3x-9}{5}=\dfrac{3-x}{2}\\ \Leftrightarrow6x-18=15-5x\\ \Leftrightarrow11x=33\\ \Leftrightarrow x=3\)
\(3,\dfrac{2x-1}{5}-\dfrac{3-x}{3}=1\\ \Leftrightarrow\dfrac{6x-3-15+5x}{15}=1\\ \Leftrightarrow11x-18=1\\ \Leftrightarrow x=\dfrac{19}{11}\)
\(4,\dfrac{x-5}{3}+\dfrac{3x+4}{2}=\dfrac{5x+2}{6}\\ \Leftrightarrow2x-10+9x+12=5x+2\\ \Leftrightarrow6x=0\Leftrightarrow x=0\)
\(5,\dfrac{x-3}{2}+\dfrac{2x+3}{5}=\dfrac{2x+5}{10}\\ \Leftrightarrow5x-15+4x+6=2x+5\\ \Leftrightarrow7x=14\\ \Leftrightarrow x=2\)
Tick nha
2: Ta có: \(\dfrac{3x-9}{5}=\dfrac{3-x}{2}\)
\(\Leftrightarrow6x-18=15-5x\)
\(\Leftrightarrow11x=33\)
hay x=3
a. \(\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)
<=> \(5\left(5x+2\right)-10\left(8x-1\right)=6\left(4x+2\right)-6\cdot5\)
<=> \(25x+10-80x+10=24x+12-30\)
<=> \(25x-80x-24x=12-30-10-10\)
<=> \(-79x=-38\)
<=> \(x=\dfrac{-38}{-79}\)
\(x=\dfrac{38}{79}\)
b. \(x-\dfrac{2x-5}{5}+\dfrac{x+8}{6}=7+\dfrac{x-1}{3}\)
<=> \(30\cdot x-6\left(2x-5\right)+5\left(x+8\right)=30\cdot7+10\left(x-1\right)\)
<=> \(30x-12x+30+5x+40=210+10x-10\)
<=> \(30x-12x+5x-10x=210-10-30-40\)
<=> \(13x=130\)
<=> \(x=\dfrac{130}{13}\)
\(x=10\)
c. \(\dfrac{x+1}{15}+\dfrac{x+2}{7}+\dfrac{x+4}{4}+6=0\)
<=> \(28\left(x+1\right)+60\left(x+2\right)+105\left(x+4\right)+420\cdot6=0\)
<=> \(28x+28+60x+120+105x+420+2520=0\)
<=> \(28x+60x+105x=-28-120-420-2520\)
<=> \(193x=-3088\)
<=> \(x=\dfrac{-3088}{193}\)
\(x=-16\)
d. \(\dfrac{x-342}{15}+\dfrac{x-323}{17}+\dfrac{x-300}{19}+\dfrac{x-273}{21}=10\)
<=> \(6783\left(x-342\right)+5985\left(x-323\right)+5355\left(x-300\right)+4845\left(x-273\right)=101745\cdot10\)
<=> \(6783x-2319786+5985x-1933155+5355x-1606500+4845x-1322685=1017450\)
<=> \(6783x+5985x+5355x+4845x=1017450+2319786+1933155+1606500+1322685\)
<=> \(22968x=8199576\)
<=> \(x=\dfrac{8199576}{22968}\)
\(x=357\)
Đề là giải PT nha các bn