\(P=15x^6-0,75x^5+2x^3-x+8\)
\(Q=x^5-3x^4+\frac{1}{2}x^3-x^2-5\)
Tính P+Q; P-Q; Q-P
bài 2
a 0,75x (x +5 )= x+5 (3- 1,25x)
b\(\frac{4}{5}x-3=\frac{1}{5}x\left(4x-15\right)\)
c(x-3) -\(\frac{\left(x-3\right)\left(2x-5\right)}{6}=\frac{\left(x-3\right)\left(3-x\right)}{4}\)
d \(\frac{\left(3x+1\right)\left(3x-2\right)}{3}+5\left(3x+1\right)=\frac{2\left(2x+1\right)\left(3x+1\right)}{3}+2x\left(3x+1\right)\)
giải phương trình sau
a) 0,75x ( x + 5 ) = ( x + 5 ) ( 3 - 1,25x )
b) \(\frac{4}{5}\)- 3 = \(\frac{1}{5}\)x ( 4x - 15 )
c) ( x - 3 ) - \(\frac{\left(x-3\right)\left(2x-5\right)}{6}\)= \(\frac{\left(x-3\right)\left(3-x\right)}{4}\)
d) \(\frac{\left(3x+1\right)\left(3x-2\right)}{3}\)+ 5( 3x + 1 ) = \(\frac{2\left(2x+1\right)\left(3x+1\right)}{3}\)+2x ( 3x +1 )
a) 0,75x(x + 5) = (x + 5)(3 - 1,25x)
<=> 0,75x(x + 5) - (x + 5)(3 - 1,25x) = (x + 5)(3 - 1,25x) - (x + 5)(3 - 1,25x)
<=> 0,75x(x + 5) - (x + 5)(3 - 1,25x) = 0
<=> (x + 5)(0,75 + 1,25x - 3) = 0
<=> (x + 5)(2x - 3) = 0
<=> x + 5 = 0 hoặc 2x - 3 = 0
<=> x = -5 hoặc x = 3/2
b) 4/5 - 3 = 1/5x(4x - 15)
<=> -11/5 = x(4x - 15)/5
<=> -11 = x(4x - 15)
<=> -11 = 4x2 - 15x
<=> 11 + 4x2 - 15x = 0
<=> 4x2 - 4x - 11x + 11 = 0
<=> 4x(x - 1) - 11(x - 1) = 0
<=> (4x - 11)(x - 1) = 0
<=> 4x - 11 = 0 hoặc x - 1 = 0
<=> x = 11/4 hoặc x = 1
c) \(\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}=\frac{\left(x-3\right)\left(3-x\right)}{4}\)
<=> 12x - 36 - 2(x - 3)(2x - 5) = 3(x - 3)(3 - x)
<=> 12x - 36 - 4x2 + 10x + 12x - 30 = 9x - 3x2 - 27 + 9x
<=> 34x - 66 - 4x2 = 18x - 3x2 - 27
<=> 34x - 66 - 4x2 - 18x + 3x2 + 27 = 0
<=> 16x - 39x - x2 = 0
<=> x2 - 16x + 39x = 0
<=> (x - 3)(x - 13) = 0
<=> x - 3 = 0 hoặc x - 13 = 0
<=> x = 3 hoặc x = 13
d) \(\frac{\left(3x+1\right)\left(3x-2\right)}{3}+5\left(3x+1\right)=\frac{2\left(2x+1\right)\left(3x+1\right)}{3}+2x\left(3x+1\right)\)
<=> (3x + 1)(3x - 2) + 15(3x + 1) = 2(2x + 1)(3x + 1) + 6x(3x + 1)
<=> 9x2 - 6x + 3x - 2 + 45x + 15 = 12x3 + 4x + 6x + 2 + 18x2 + 6x
<=> 9x2 + 42x + 13 = 30x2 + 16x + 2
<=> 9x2 + 42x + 13 - 30x2 - 16x - 2 = 0
<=> -21x2 + 26x + 11 = 0
<=> 21x2 - 26x - 11 = 0
<=> 21x2 + 7x - 33x - 11 = 0
<=> 7x(3x + 1) - 11(3x + 1) = 0
<=> (7x - 11)(3x + 1) = 0
<=> 7x - 11 = 0 hoặc 3x + 1 = 0
<=> x = 11/7 hoặc x = -1/3
Giải phương trình:
1. \(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
2. \(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
3. \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{15}\)
4. \(\frac{x+3}{2}-\frac{x-1}{3}=\frac{x+5}{6}+1\)
5. \(\frac{x-4}{5}-\frac{3x-2}{10}-x=\frac{2x-5}{3}-\frac{7x+2}{6}\)
6. \(\frac{\left(x+2\right)\left(x+10\right)}{3}-\frac{\left(x+4\right)\left(x+10\right)}{12}=\frac{\left(x-2\right)\left(x+4\right)}{4}\)
7. \(\frac{\left(x+2\right)^2}{8}-2\left(2x-1\right)=25+\frac{\left(x-2\right)^2}{8}\)
8.\(\frac{7x^2-14x-5}{5}=\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}\)
9. \(\frac{\left(2x-3\right)\left(2x+3\right)}{8}=\frac{\left(x-4\right)^2}{6}+\frac{\left(x-2\right)^2}{3}\)
10. \(\frac{x+1}{35}+\frac{x+3}{33}=\frac{x+5}{31}+\frac{x+7}{29}\)
1.
\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
\(MC:12\)
Quy đồng :
\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)
\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)
\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)
\(\Leftrightarrow6x+9-3x=-4-9+16\)
\(\Leftrightarrow-7x=3\)
\(\Leftrightarrow x=\frac{-3}{7}\)
2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
\(MC:20\)
Quy đồng :
\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)
\(\Leftrightarrow30x+15-20=15x-2\)
\(\Leftrightarrow15x=3\)
\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)
Giải pt:
\(a.\frac{x}{3}-\frac{5x}{6}-\frac{15x}{12}=\frac{x}{4}-5\)
\(b.\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
\(c.\frac{x-1}{2}-\frac{x+1}{15}-\frac{2x-13}{6}=0\)
\(d.\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}=\frac{1-x}{2}-2\)
a) \(\frac{x}{3}-\frac{5x}{6}-\frac{15x}{12}=\frac{x}{4}-5\)
\(\Leftrightarrow\frac{4x-10x-15x}{12}=\frac{3x-60}{12}\)
\(\Leftrightarrow-21x=3x-60\)
\(\Leftrightarrow24x=60\)
\(\Leftrightarrow x=\frac{5}{2}\)
Vậy tập nghiệm của phương trình là \(S=\left\{\frac{5}{2}\right\}\)
b) \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
\(\Leftrightarrow\frac{\left(8x-3\right)-2\left(3x-2\right)}{4}=\frac{2\left(2x-1\right)+\left(x+3\right)}{4}\)
\(\Leftrightarrow8x-3-6x+4=4x-2+x+3\)
\(\Leftrightarrow2x+1=5x+1\)
\(\Leftrightarrow2x=5x\)
\(\Leftrightarrow x=0\)
Vậy tập nghiệm của phương trình là \(S=\left\{0\right\}\)
c) \(\frac{x-1}{2}-\frac{x+1}{15}-\frac{2x-13}{6}=0\)
\(\Leftrightarrow\frac{15\left(x-1\right)-2\left(x+1\right)-5\left(2x-13\right)}{30}=0\)
\(\Leftrightarrow15x-15-2x-2-10x+65=0\)
\(\Leftrightarrow3x+48=0\)
\(\Leftrightarrow x=-16\)
Vậy tập nghiệm của phương trình là \(S=\left\{-16\right\}\)
d) \(\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}=\frac{1-x}{2}-2\)
\(\Leftrightarrow\frac{9\left(3-x\right)+16\left(5-x\right)}{24}=\frac{12\left(1-x\right)-48}{24}\)
\(\Leftrightarrow27-9x+80-16x=12-12x-48\)
\(\Leftrightarrow-25x+107=-12x-36\)
\(\Leftrightarrow-13x+143=0\)
\(\Leftrightarrow x=11\)
Vậy tập nghiệm của phương trình là \(S=\left\{11\right\}\)
Bài 2. Giải các phương trình sau
a, \(\frac{x}{3}-\frac{5x}{6}-\frac{15x}{12}=\frac{x}{4}-5\)
b, \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
c, \(\frac{x-1}{2}-\frac{x+1}{15}-\frac{2x-13}{6}=0\)
d,\(\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}=\frac{1-x}{2}-2\)
e, \(\frac{3\left(5x-2\right)}{4}-2=\frac{7x}{3}-5\left(x-7\right)\)
f, \(\frac{x+5}{2}+\frac{3-2x}{4}=x-\frac{7+x}{6}\)
g, \(\frac{x-3}{11}+\frac{x+1}{3}=\frac{x+7}{9}-1\)
h, \(\frac{3x-0,4}{2}+\frac{1,5-2x}{3}=\frac{x+0,5}{5}\)
a)
\(\frac{x}{3}-\frac{5x}{6}-\frac{15x}{12}=\frac{x}{4}-5\)
\(\Leftrightarrow\frac{4x-10x-15x}{12}=\frac{3x-60}{12}\)
\(\Leftrightarrow\frac{-10x-11}{12}=\frac{3x-60}{12}\)
\(\Leftrightarrow\frac{-10x-11-3x+60}{12}=0\)
\(\Leftrightarrow\frac{49-13x}{12}=0\)
\(\Rightarrow49-13x=0\)
\(\Rightarrow x=\frac{-49}{13}\)
b)
\(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
\(\Leftrightarrow\frac{8x-3-6x+4}{4}=\frac{4x-2+x+3}{4}\)
\(\Leftrightarrow\frac{2x+1}{4}=\frac{5x+1}{4}\)
\(\Leftrightarrow\frac{2x+1-5x-1}{4}=0\)
\(\Leftrightarrow\frac{-3x}{4}=0\)
\(\Rightarrow-3x=0\)
\(\Rightarrow x=0\)
e)
\(\frac{3\cdot\left(5x-2\right)}{4}-2=\frac{7x}{3}-5\cdot\left(x-7\right)\)
\(\Leftrightarrow\frac{45x-18-24-28x+60x-420}{12}=0\)
\(\Leftrightarrow\frac{77x-462}{12}=0\)
\(\Rightarrow77x-462=0\)
\(\Rightarrow x=\frac{462}{77}=6\)
Câu 3: Giải các phương trình sau bằng cách đưa về dạng ax+b=0
1. a, \(\frac{5x-2}{3}=\frac{5-3x}{2}\); b, \(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)
c, \(2\left(x+\frac{3}{5}\right)=5-\left(\frac{13}{5}+x\right)\); d, \(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\)
e, \(\frac{7x-1}{6}+2x=\frac{16-x}{5}\); f, 4 (0,5-1,5x)=\(\frac{5x-6}{3}\)
g, \(\frac{3x+2}{2}-\frac{3x+1}{6}=\frac{5}{3}+2x\); h, \(\frac{x+4}{5}.x+4=\frac{x}{3}-\frac{x-2}{2}\)
i, \(\frac{4x+3}{5}-\frac{6x-2}{7}=\frac{5x+4}{3}+3\); k, \(\frac{5x+2}{6}-\frac{8x-1}{3}=\frac{4x+2}{5}-5\)
m, \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{15}\); n, \(\frac{1}{4}\left(x+3\right)=3-\frac{1}{2}\left(x+1\right).\frac{1}{3}\left(x+2\right)\)
p, \(\frac{x}{3}-\frac{2x+1}{6}=\frac{x}{6}-x\); q, \(\frac{2+x}{5}-0,5x=\frac{1-2x}{4}+0,25\)
r, \(\frac{3x-11}{11}-\frac{x}{3}=\frac{3x-5}{7}-\frac{5x-3}{9}\); s, \(\frac{9x-0,7}{4}-\frac{5x-1,5}{7}=\frac{7x-1,1}{6}-\frac{5\left(0,4-2x\right)}{6}\)
t, \(\frac{2x-8}{6}.\frac{3x+1}{4}=\frac{9x-2}{8}+\frac{3x-1}{12}\); u, \(\frac{x+5}{4}-\frac{2x-3}{3}=\frac{6x-1}{3}+\frac{2x-1}{12}\)
v, \(\frac{5x-1}{10}+\frac{2x+3}{6}=\frac{x-8}{15}-\frac{x}{30}\); w, \(\frac{2x-\frac{4-3x}{5}}{15}=\frac{7x\frac{x-3}{2}}{5}-x+1\)
Đây là những bài cơ bản mà bạn!
\(\frac{5x-2}{3}=\frac{5-3x}{2}\)
\(< =>\frac{\left(5x-2\right).2}{6}=\frac{\left(5-3x\right).3}{6}\)
\(< =>\left(5x-2\right).2=\left(5-3x\right).3\)
\(< =>10x-4=15-9x\)
\(< =>10x+9x=15+4\)
\(< =>19x=19< =>x=1\)
\(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)
\(< =>\frac{\left(10x+3\right).3}{36}=\frac{36}{36}+\frac{\left(6+8x\right).4}{36}\)
\(< =>\left(10x+3\right).3=36+\left(6+8x\right).4\)
\(< =>30x+9=36+24+32x\)
\(< =>32x-30x=9-36-24\)
\(< =>2x=9-60=-51< =>x=-\frac{51}{2}\)
giải các hệ BPT sau:
a) \(\left\{{}\begin{matrix}5x-2>4x+5\\5x-4< x+2\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}2x+1>3x+4\\5x+3\ge8x-9\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}\frac{5x+2}{3}\ge4-x\\\frac{6-5x}{13}< 3x+1\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}\frac{4x-5}{7}< x+3\\\frac{3x+8}{4}>2x-5\end{matrix}\right.\)
e) \(\left\{{}\begin{matrix}6x+\frac{5}{7}< 4x+7\\\frac{8x+3}{2}< 2x+5\end{matrix}\right.\)
f) \(\left\{{}\begin{matrix}15x-2>2x+\frac{1}{3}\\2\left(x-4\right)< \frac{3x-14}{2}\end{matrix}\right.\)
g) \(\left\{{}\begin{matrix}x-1\le2x-3\\3x< x+5\\5-3x\le2x-6\end{matrix}\right.\)
h) \(\left\{{}\begin{matrix}2x+\frac{3}{5}>\frac{3\left(2x-7\right)}{3}\\x-\frac{1}{2}< \frac{5\left(3x-1\right)}{2}\end{matrix}\right.\)
j) \(\left\{{}\begin{matrix}\frac{3x+1}{2}-\frac{3-x}{3}\le\frac{x+1}{4}-\frac{2x-1}{3}\\3-\frac{2x+1}{5}>x+\frac{4}{3}\end{matrix}\right.\)
bài 1
a 0,75x(x+5)=x+5(3-1,25x)
b \(\frac{4}{5}x-3=\frac{1}{5}x\left(4x-15\right)\)
c \(\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}=\frac{\left(x-3\right)\left(3x-x\right)}{4}\)
Giải phương trình:
2x-2=8-3x
x2-3x+1=x+x2
(x2+1)(2x+4)=0
(4x+1)(x2+2)=0
\(\frac{x}{2}=3-\frac{x+4}{3}\)
\(\frac{3-x}{4}=1-\frac{3x-5}{6}\)
\(\frac{2x+5}{9}=2+\frac{x-3}{6}\)
\(\frac{x+5}{3}=1+\frac{x-3}{9}\)
\(\frac{2x-5}{x+5}=3\)
\(\frac{x^2-6}{x}=x+\frac{3}{2}\)
\(\frac{\left(x^2+2x\right)-\left(3x+6\right)}{x-3}=0\)
\(\frac{5}{3x+2}=2x-1\)
\(2x-2=8-3x\)
\(\Leftrightarrow\)\(2x+3x=8+2\)
\(\Leftrightarrow\)\(5x=10\)
\(\Leftrightarrow\)\(x=2\)
Vậy...
\(x^2-3x+1=x+x^2\)
\(\Leftrightarrow\)\(x^2-3x-x-x^2=-1\)
\(\Leftrightarrow\)\(-4x=-1\)
\(\Leftrightarrow\)\(x=\frac{1}{4}\)
Vậy...
mấy cái này bấm máy tính là đc òi. giải mất thời gian lắm :))
\(2x-2=8-3x\)
\(\Leftrightarrow\)\(2x+3x=8+2\)
\(\Leftrightarrow\)\(5x=10\)
\(\Leftrightarrow\)\(x=2\)
Vậy \(x=2\)
\(x^2-3x+1=x+x^2\)
\(\Leftrightarrow\)\(4x-1=\left(x^2+x\right)-\left(x^2+x\right)\)
\(\Leftrightarrow\)\(4x-1=0\)
\(\Leftrightarrow\)\(4x=1\)
\(\Leftrightarrow\)\(x=\frac{1}{4}\)
Vậy \(x=\frac{1}{4}\)
\(\left(x^2+1\right)\left(2x+4\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x^2+1=0\\2x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2=-1\\2x=-4\end{cases}}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\sqrt{-1}\\x=\frac{-4}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x\in\left\{\varnothing\right\}\\x=-2\end{cases}}}\)
Vậy \(x=-2\)
Chúc bạn học tốt ~