Câu 1:Tìm nghiệm của các pt sau:
a) x - 1 = 5 - x
b) 3+ x = 2 - x
Câu 2:giải pt sau:
a) 3x + 7 = 2x - 3
b) 4 - (x - 2) = (3 - 2x)
Câu 3:giải các pt sau:
a.\(\dfrac{5x-4}{2}\)=\(\dfrac{16x+1}{7}\)
b.\(\dfrac{12x+5}{3}\)=\(\dfrac{2x-7}{4}\)
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
giải pt :
a, (x+5)(2-x)=3\(\sqrt{x^2+3x}\)
b, \(\sqrt[3]{\dfrac{2x}{x+1}}+\sqrt[3]{\dfrac{1}{2}+\dfrac{1}{2x}}=2\)
c,\(\sqrt[5]{\dfrac{16x}{x-1}}+\sqrt[5]{\dfrac{x-1}{16x}}=\dfrac{5}{2}\)
d, \(\sqrt{5x^2+10x+1}=7-2x-x^2\)
e, \(\sqrt{2x^2+4x+1}=1-2x-x^2\)
GIẢI CÁC PT SAU:
\(\dfrac{2x+1}{3x+2}=5\)
\(\dfrac{2x^2-5x+2}{x-1}=\dfrac{2x^2+x+15}{x-3}\)
\(\dfrac{2x+3}{x-3}-\dfrac{4}{x+3}=\dfrac{24}{x^2-9}+2\)
Bài 1:
a) Giải PT sau: \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)
b) Giải PT sau: |2x+6|-x=3
a) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-4}{\left(x-2\right)\left(x+2\right)}\)
Suy ra: \(x^2+3x+2-5x+10=12+x^2-4\)
\(\Leftrightarrow x^2-2x+12-8-x^2=0\)
\(\Leftrightarrow-2x+4=0\)
\(\Leftrightarrow-2x=-4\)
hay x=2(loại)
Vậy: \(S=\varnothing\)
b) Ta có: \(\left|2x+6\right|-x=3\)
\(\Leftrightarrow\left|2x+6\right|=x+3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+6=x+3\left(x\ge-3\right)\\-2x-6=x+3\left(x< -3\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-x=3-6\\-2x-x=3+6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=-3\left(loại\right)\end{matrix}\right.\)
Vậy: S={-3}
câu 1 giải các pt sau
a,3x-12=0 b,(x-2)(2x+3)=0 c,\(\dfrac{x+2}{x-2}-\dfrac{6}{x+2}=\dfrac{x^2}{x^2-4}\)
\(a,3x-12=0\)
\(\Leftrightarrow3x=12\)
\(\Leftrightarrow x=4\)
\(b,\left(x-2\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(c,\dfrac{x+2}{x-2}-\dfrac{6}{x+2}=\dfrac{x^2}{x^2-4}\left(dkxd:x\ne\pm2\right)\)
\(\Leftrightarrow\dfrac{\left(x+2\right)^2-6\left(x-2\right)-x^2}{x^2-4}=0\)
\(\Leftrightarrow x^2+4x+4-6x+12-x^2=0\)
\(\Leftrightarrow-2x+16=0\)
\(\Leftrightarrow-2x=-16\)
\(\Leftrightarrow x=8\left(tmdk\right)\)
\(a,3x-12=0\)
\(\Leftrightarrow3x=12\)
\(\Leftrightarrow x=4.\)
Vậy \(S=\left\{4\right\}\)
\(b,\left(x-2\right)\left(2x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\2x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2.\\x=\dfrac{-3}{2}.\end{matrix}\right.\)
Vậy \(S=\left\{2;\dfrac{-3}{2}\right\}\)
\(c,\dfrac{x+2}{x-2}-\dfrac{6}{x+2}=\dfrac{x^2}{x^2-4}\left(ĐKXĐ:x\ne\pm2\right)\)
\(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{6\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{x^2}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\dfrac{x^2+4x+4}{\left(x-2\right)\left(x+2\right)}-\dfrac{6x-12}{\left(x-2\right)\left(x+2\right)}-\dfrac{x^2}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Rightarrow x^2+4x+4-6x+12-x^2=0\)
\(\Leftrightarrow-2x+16=0\)
\(\Leftrightarrow-2x=-16\)
\(\Leftrightarrow x=8\left(tm\right).\)
Vậy \(S=\left\{8\right\}\)
Giải các pt sau:
1)\(\dfrac{2x+1}{x^2-4}+\dfrac{2}{x+1}=\dfrac{3}{2-x}\)
2)\(\dfrac{3x+1}{1-3x}+\dfrac{3+x}{3-x}=2\)
3)\(\dfrac{8x-2}{3}=1+\dfrac{5-2x}{4}\)
4)
\(\dfrac{x}{x+1}-\dfrac{2x+3}{x}=\dfrac{-3}{x+1}-\dfrac{3}{x}\)
5)\(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{x^2-1}\)
6)\(\dfrac{2x+5}{2x}-\dfrac{x}{x+5}=0\)
giúp mình với cám ơn
1: Sửa đề: 2/x+2
\(\dfrac{2x+1}{x^2-4}+\dfrac{2}{x+2}=\dfrac{3}{2-x}\)
=>\(\dfrac{2x+1+2x-4}{x^2-4}=\dfrac{-3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
=>4x-3=-3x-6
=>7x=-3
=>x=-3/7(nhận)
2: \(\Leftrightarrow\dfrac{\left(3x+1\right)\left(3-x\right)+\left(3+x\right)\left(1-3x\right)}{\left(1-3x\right)\left(3-x\right)}=2\)
=>9x-3x^2+3-x+3-9x+x-3x^2=2(3x-1)(x-3)
=>-6x^2+6=2(3x^2-10x+3)
=>-6x^2+6=6x^2-20x+6
=>-12x^2+20x=0
=>-4x(3x-5)=0
=>x=5/3(nhận) hoặc x=0(nhận)
3: \(\Leftrightarrow x\cdot\dfrac{8}{3}-\dfrac{2}{3}=1+\dfrac{5}{4}-\dfrac{1}{2}x\)
=>x*19/6=35/12
=>x=35/38
Mấy bạnn giải chii tiết raa giúp mik với nhaa Câu 1: nghiệm dương nhỏ nhất của pt tan x=tan (6π/5) A. x=π/5 B. x=6π/5 C. x=6/5 D. x=6π Câu 2: tìm nghiệm thuộc đoạn [0;π] của pt cot 2x=cot(π/2-x) A. 2 B. 3 C.1 D.4 Câu 3: tìm tổng các nghiệm thuộc khoảng (-π/2;π/2) của pt 4sin²2x-1=0 A.0 B. π/6 C. π/3 D. π Câu 4: tìm tổng các nghiệm của pt cos(x+π/4)=1/2 trong khoảng (-π;π) A. π/2 B. -π/2 C. -3π/2 D. π/4
Bài 1 : Giải các pt sau :
c) |2x - 1| = x + 2
Bài 2 : giải các BPT sau :
a) 2( 3x - 1 ) < x + 4
b) 5 -2x/3 + x ≥ x/2 + 1
Bài 1:
c) |2x - 1| = x + 2
<=> 2x - 1 = +(x + 2) hoặc -(x + 2)
* 2x - 1 = x + 2
<=> 2x - x = 2 + 1
<=> x = 3
* 2x - 1 = -(x + 2)
<=> 2x - 1 = x - 2
<=> 2x - x = -2 + 1
<=> x = -1
Vậy.....