giải các pt sau
a,2x(x-7)-(x-1)2 -x2 +7
b,x(x+3)2 -3x = (x+2)3 +1
c, 2(x-1)2 +(x+3)3=3(x-2)(x+1)
câu 1:giải các pt và bpt sau: a,17x - 5(x+3)= 2x + 5 b,3/x+2 - 5/x-2 = 11x + 23/(x+2)(x-2) c,5x + 7 ≥ 3(x-1) d,3x-1/x+1 = -2/5 e,(2x-1)(2x+1)= 4x2 + 3x + 2 f,x-3^3 -7+3x g,7x-5 < 2(4x-1)+7
a: =>17x-5x-15-2x-5=0
=>10x-20=0
=>x=2
b: =>\(\dfrac{3x-6-5x-10}{\left(x+2\right)\left(x-2\right)}=\dfrac{11x+23}{\left(x+2\right)\left(x-2\right)}\)
=>11x+23=-2x-16
=>13x=-39
=>x=-3(nhận)
c: =>5x+7>=3x-3
=>2x>=-10
=>x>=-5
d: =>5(3x-1)=-2(x+1)
=>15x-5=-2x-2
=>17x=3
=>x=3/17
e: =>4x^2-1-4x^2-3x-2=0
=>-3x-3=0
=>x=-1
g: =>7x-5-8x+2-7<0
=>-x-10<0
=>x+10>0
=>x>-10
câu 1:giải các pt và bpt sau:
a,17x - 5(x+3)= 2x + 5
b,3/x+2 - 5/x-2 = 11x + 23/(x+2)(x-2)
c,5x + 7 ≥ 3(x-1)
d,3x-1/x+1 = -2/5
e,(2x-1)(2x+1)= 4x2 + 3x + 2
f,x-3^3 -7+3x
g,7x-5 < 2(4x-1)+7
a: =>17x-5x-15-2x-5=0
=>10x-20=0
=>x=2
b: =>\(\dfrac{3x-6-5x-10}{\left(x+2\right)\left(x-2\right)}=\dfrac{11x+23}{\left(x+2\right)\left(x-2\right)}\)
=>11x+23=-2x-16
=>13x=-39
=>x=-3(nhận)
c: =>5x+7>=3x-3
=>2x>=-10
=>x>=-5
d: =>5(3x-1)=-2(x+1)
=>15x-5=-2x-2
=>17x=3
=>x=3/17
e: =>4x^2-1-4x^2-3x-2=0
=>-3x-3=0
=>x=-1
g: =>7x-5-8x+2-7<0
=>-x-10<0
=>x+10>0
=>x>-10
đang gấp, giúp e với ạ
bài 1: giải các pt sau: a, 1/x-2+3=7-x/x-2
b, x^2-x/x+3-x^2/x-3=7x^2-3x/9-x^2
c, x/2x-1 + x+1/x+2=0
d, 2/x+1 - 1/x-2 = 3x-11/(x+1)(x-2)
a: =>1+3x-6=7-x
=>3x-5=7-x
=>4x=12
=>x=3(nhận)
b: \(\Leftrightarrow\dfrac{x^2-x}{x+3}-\dfrac{x^2}{x-3}=\dfrac{-7x^2+3x}{\left(x-3\right)\left(x+3\right)}\)
=>\(x^3-3x^2-x^2+3x-x^3-3x^2=-7x^2+3x\)
=>\(-7x^2+3x=-7x^2+3x\)
=>0x=0(luôn đúng)
Vậy: S=R\{3;-3}
c: =>x(x+2)+(2x-1)(x+1)=0
=>2x^2+2x-x-1+x^2+2x=0
=>3x^2+3x-1=0
\(x=\dfrac{-3\pm\sqrt{21}}{6}\)
d: =>2(x-2)-x-1=3x-11
=>3x-11=2x-4-x-1=x-5
=>2x=6
=>x=3(nhận)
Bài 1. Giải các phương trình sau bằng cách đưa về dạng ax + b = 0:
1. a) 5 – (x – 6) = 4(3 – 2x) b) 2x(x + 2)2 – 8x2 = 2(x – 2)(x2 + 2x + 4)
c) 7 – (2x + 4) = – (x + 4) d) (x – 2)3 + (3x – 1)(3x + 1) = (x + 1)3
e) (x + 1)(2x – 3) = (2x – 1)(x + 5) f) (x – 1)3 – x(x + 1)2 = 5x(2 – x) – 11(x + 2)
g) (x – 1) – (2x – 1) = 9 – x h) (x – 3)(x + 4) – 2(3x – 2) = (x – 4)2
i) x(x + 3)2 – 3x = (x + 2)3 + 1 j) (x + 1)(x2 – x + 1) – 2x = x(x + 1)(x – 1)
2. a) b)
c) d)
e) f)
g) h)
i) k)
m) n)
bạn đăng tách cho mn cùng giúp nhé
Bài 1 :
a, \(\Leftrightarrow11-x=12-8x\Leftrightarrow7x=1\Leftrightarrow x=\dfrac{1}{7}\)
b, \(\Leftrightarrow2x\left(x^2+4x+4\right)-8x^2=2\left(x^3-8\right)\)
\(\Leftrightarrow2x^3+8x^2+8x-8x^2=2x^3-16\Leftrightarrow x=-2\)
c, \(\Leftrightarrow3-2x=-x-4\Leftrightarrow x=7\)
d, \(\Leftrightarrow x^3-6x^2+12x-8+9x^2-1=x^3+3x^2+3x+1\)
\(\Leftrightarrow3x^2+12x-9=3x^2+3x+1\Leftrightarrow x=\dfrac{10}{9}\)
e, \(\Leftrightarrow2x^2-x-3=2x^2+9x-5\Leftrightarrow x=5\)
f, \(\Leftrightarrow x^3-3x^2+3x-1-x^3-2x^2-x=10x-5x^2-11x-22\)
\(\Leftrightarrow-5x^2+2x-1=-5x^2-x-22\Leftrightarrow3x=-21\Leftrightarrow x=-7\)
h) \(PT\Leftrightarrow x^2+4x-3x-12-6x+4=x^2-8x+16\)
\(\Leftrightarrow3x=24\)
\(\Leftrightarrow x=8\)
Vậy: \(S=\left\{8\right\}\)
j) \(PT\Leftrightarrow x^3-x^2+x+x^2-x+1-2x=x^3-x\)
\(\Leftrightarrow x=1\)
Vậy: \(S=\left\{1\right\}\)
1.giải các pt sau
a)3x-7/2+x-1/3=-16
b)x-x-1/3=2x+1/5
c)2x-1/3-5x+2/7=x+13
d)2x-3/3-x-3/6=4x+3/5-17
a: \(\dfrac{3x-7}{2}+\dfrac{x-1}{3}=-16\)
\(\Leftrightarrow3\left(3x-7\right)+2\left(x-1\right)=-96\)
\(\Leftrightarrow9x-21+2x-2=-96\)
=>11x=-73
hay x=-73/11
b: \(x-\dfrac{x-1}{3}=\dfrac{2x+1}{5}\)
=>15x-5(x-1)=3(2x+1)
=>15x-5x+5=6x+3
=>10x+5=6x+3
=>4x=-2
hay x=-1/2
c: \(\dfrac{2x-1}{3}-\dfrac{5x+2}{7}=x+13\)
=>14x-7-15x-6=21(x+13)
=>21x+273=-x-13
=>22x=-286
hay x=13
Giải các pt sau:
a) (x-3)-(x-3)(2x-5)/6=(x-3)(3-x)/4
b) (2x-7)^2-x^2+8x-16=0
c) (3x+1)(x-3)=(3x+1)(2x-5)
\(\left(3x+1\right)\left(x-3\right)=\left(3x+1\right)\left(2x-5\right)\)
\(\Leftrightarrow\left(3x+1\right)\left(x-3\right)-\left(3x+1\right)\left(2x-5\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(x-3-2x+5\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(2-x\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}3x+1=0\\2-x=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[\begin{matrix}3x=-1\\x=2\end{matrix}\right.\)
\(\Leftrightarrow\left[\begin{matrix}x=-\frac{1}{3}\\x=2\end{matrix}\right.\)
Vậy tập nghiệm của pt là \(S=\left\{-\frac{1}{3};2\right\}\)
Có : \(\left(3x+1\right)\left(x-3\right)=\left(3x+1\right)\left(2x-5\right)\)
\(\Leftrightarrow\) \(\left(3x+1\right)\left(x-3\right)-\left(3x+1\right)\left(2x-5\right)=0\)
\(\Leftrightarrow\) \(\left(3x+1\right)\left(x-3-2x+5\right)=0\)
\(\Leftrightarrow\) \(\left(3x+1\right)\left(-x+2\right)=0\)
\(\Leftrightarrow\) \(\left[\begin{matrix}3x+1=0\\-x+2=0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left[\begin{matrix}3x=-1\\-x=-2\end{matrix}\right.\) \(\Leftrightarrow\) \(\left[\begin{matrix}x=\frac{-1}{3}\\x=2\end{matrix}\right.\)
Vậy phương trình có tập nghiệm \(S=\left\{\frac{-1}{3};2\right\}\)
\(\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}=\frac{\left(x-3\right)\left(3-x\right)}{4}\)
\(\Leftrightarrow\frac{24\left(x-3\right)}{24}-\frac{4\left(x-3\right)\left(2x-5\right)}{24}=-\frac{6\left(x-3\right)\left(x-3\right)}{24}\)
\(\Leftrightarrow24\left(x-3\right)-4\left(x-3\right)\left(2x-5\right)+6\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(x-3\right)\left[24-4\left(2x-5\right)+6\left(x-3\right)\right]=0\)
\(\Leftrightarrow\left(x-3\right)\left(24-8x+20+6x-18\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(26-2x\right)=0\)
\(\Leftrightarrow2\left(x-3\right)\left(13-x\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x-3=0\\13-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[\begin{matrix}x=3\\x=13\end{matrix}\right.\)
Vậy tập nghiệm của pt là \(S=\left\{3;13\right\}\)
Bài1: giải các pt sau:
a, 3-4x+24+6x= x+27+3x
b, 5-(6-x)=4(3-2x)
c, x-(x+1)/3 = (2x+1)/5
d,(2x-1)/3 - (5x+2)/7 = x+13
Bài 2:
a, (x-1)(3x+1)=0
b, (x-5)(7-x)=0
c, ( x-1)(x+5)(-3x+8)=0
d, x(x^2 - 1 )=0
Giúp mình 2 bài này với , mình đang cần gấp , CẢM ƠN M.N ạ><
2:
a: =>x-1=0 hoặc 3x+1=0
=>x=1 hoặc x=-1/3
b: =>x-5=0 hoặc 7-x=0
=>x=5 hoặc x=7
c: =>\(\left[{}\begin{matrix}x-1=0\\x+5=0\\3x-8=0\end{matrix}\right.\Leftrightarrow x\in\left\{1;-5;\dfrac{8}{3}\right\}\)
d: =>x=0 hoặc x^2-1=0
=>\(x\in\left\{0;1;-1\right\}\)
1.giải các phương trình sau:
a, 3(2x+1)/4 - 5x+3/6 = 2x-1/3 - 3-x/4
b, 19/4 - 2(3x-5)/5 = 3-2x/10 - 3x-1/4
c, x-2*3/2+3 + x-3*5/3+5 + x-5*2/5+2 = 10
d, x-3/5*7 + x-5/3*7 + x-7/3*5 = 2(1/3 + 1/5 + 1/7)
2. giải các phương trình:
a, x-1/9 + x-2/8 = x-3/7 + x-4/6
b, (1/1*2 + 1/2*3 + 1/3*4 + ... + 1/9*10) (x-1) + 1/10x = x- 9/10
Câu 1 :
a, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}=\frac{2x-1}{3}-\frac{3-x}{4}\)
\(\Leftrightarrow\frac{6x+3}{4}+\frac{3-x}{4}=\frac{2x-1}{3}+\frac{5x+3}{6}\)
\(\Leftrightarrow\frac{5x+6}{4}=\frac{9x+1}{6}\Leftrightarrow\frac{30x+36}{24}=\frac{36x+4}{24}\)
Khử mẫu : \(30x+36=36x+4\Leftrightarrow-6x=-32\Leftrightarrow x=\frac{32}{6}=\frac{16}{3}\)
tương tự
\(\frac{19}{4}-\frac{2\left(3x-5\right)}{5}=\frac{3-2x}{10}-\frac{3x-1}{4}\)
\(< =>\frac{19.5}{20}-\frac{8\left(3x-5\right)}{20}=\frac{2\left(3-2x\right)}{20}-\frac{5\left(3x-1\right)}{20}\)
\(< =>95-24x+40=6-4x-15x+5\)
\(< =>-24x+135=-19x+11\)
\(< =>5x=135-11=124\)
\(< =>x=\frac{124}{5}\)
\(\frac{\left(x-2\right).3}{2}+3+\frac{\left(x-3\right).5}{3}+5+\frac{\left(x-5\right).2}{5}+2=10\)
\(< =>\frac{\left(x-2\right).3.15}{30}+\frac{\left(x-3\right).5.10}{30}+\frac{\left(x-5\right).2.6}{30}=10-2-3-5\)
\(< =>\frac{\left(x-2\right).45+\left(x-3\right).50+\left(x-5\right).12}{30}=0\)
\(< =>45x-90+50x-150+12x-60=0\)
\(< =>107x-300=0< =>x=\frac{300}{107}\)
Bài 1 giải các pt sau và diễn tập nghiệm trên trục số a) 2x-6>0 b) -3x+9>0 c)3(x-1)+5>(x+1)+3 d)x/3 - 1/2>x/6 Bài 2:a)cho a>b chứng minh 3a+7>3b+7 b)cho a >b chứng minh a+3>b+1 c) cho 5a -1>5b-1 hãy so sánh a và b Bài 3: 2x(x+5)=0 b) X^2-4=0 d) (x-5)(2x+1)+(x-5)(x+6)=0 Ở bài 1 câu a có dấu hoặc bằng nữa nha bài 2 câu c cũng vậy
3:
a: =>x=0 hoặc x+5=0
=>x=0 hoặc x=-5
b: =>x^2=4
=>x=2 hoặc x=-2
c: =>(x-5)(2x+1+x+6)=0
=>(x-5)(3x+7)=0
=>x=5 hoặc x=-7/3
1.
a. 2x - 6 > 0
\(\Leftrightarrow\) 2x > 6
\(\Leftrightarrow\) x > 3
S = \(\left\{x\uparrow x>3\right\}\)
b. -3x + 9 > 0
\(\Leftrightarrow\) - 3x > - 9
\(\Leftrightarrow\) x < 3
S = \(\left\{x\uparrow x< 3\right\}\)
c. 3(x - 1) + 5 > (x - 1) + 3
\(\Leftrightarrow\) 3x - 3 + 5 > x - 1 + 3
\(\Leftrightarrow\) 3x - 3 + 5 - x + 1 - 3 > 0
\(\Leftrightarrow\) 2x > 0
\(\Leftrightarrow\) x > 0
S = \(\left\{x\uparrow x>0\right\}\)
d. \(\dfrac{x}{3}-\dfrac{1}{2}>\dfrac{x}{6}\)
\(\Leftrightarrow\dfrac{2x}{6}-\dfrac{3}{6}>\dfrac{x}{6}\)
\(\Leftrightarrow2x-3>x\)
\(\Leftrightarrow2x-3-x>0\)
\(\Leftrightarrow x-3>0\)
\(\Leftrightarrow x>3\)
\(S=\left\{x\uparrow x>3\right\}\)
2.
a.
Ta có: a > b
3a > 3b (nhân cả 2 vế cho 3)
3a + 7 > 3b + 7 (cộng cả 2 vế cho 7)
b. Ta có: a > b
a > b (nhân cả 2 vế cho 1)
a + 3 > b + 3 (cộng cả 2 vế cho 3) (1)
Ta có; 3 > 1
b + 3 > b + 1 (nhân cả 2 vế cho 1b) (2)
Từ (1) và (2) \(\Rightarrow\) a + 3 > b + 1
c.
5a - 1 + 1 > 5b - 1 + 1 (cộng cả 2 vế cho 1)
5a . \(\dfrac{1}{5}\) > 5b . \(\dfrac{1}{5}\) (nhân cả 2 vế cho \(\dfrac{1}{5}\) )
a > b
3.
a. 2x(x + 5) = 0
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x+5=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
\(S=\left\{0,-5\right\}\)
b. x2 - 4 = 0
\(\Leftrightarrow x\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
\(S=\left\{0,4\right\}\)
d. (x - 5)(2x + 1) + (x - 5)(x + 6) = 0
\(\Leftrightarrow\left(x-5\right)\left(2x+1+x+6\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(3x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{-7}{3}\end{matrix}\right.\)
\(S=\left\{5,\dfrac{-7}{3}\right\}\)