1:

a: =>(|x|+4)(|x|-1)=0

=>|x|-1=0

=>x=1; x=-1

b: =>x^2-4>=0

=>x>=2 hoặc x<=-2

d: =>|2x+5|=2x-5

=>x>=5/2 và (2x+5-2x+5)(2x+5+2x-5)=0

=>x=0(loại)

a: ĐKXĐ: x∈R

\(\frac{5}{x^2-2x+2}-\frac{8}{x^2-2x+5}=3\)

=>\(\frac{5\left(x^2-2x+5\right)-8\left(x^2-2x+2\right)}{\left(x^2-2x+2\right)\left(x^2-2x+5\right)}=3\)

=>\(3\left(x^2-2x+2\right)\left(x^2-2x+5\right)=5x^2-10x+25-8x^2+16x-16=-3x^2+6x+9\)

=>\(3\left\lbrack\left(x^2-2x\right)^2+7\left(x^2-2x\right)+10\right\rbrack=-3\left(x^2-2x\right)+9\)

=>\(\left(x^2-2x\right)^2+7\left(x^2-2x\right)+10=-\left(x^2-2x\right)+3\)

=>\(\left(x^2-2x\right)^2+8\left(x^2-2x\right)+7=0\)

=>\(\left(x^2-2x+1\right)\left(x^2-2x+7\right)=0\)

=>\(x^2-2x+1=0\)

=>\(\left(x-1\right)^2=0\)

=>x-1=0

=>x=1(nhận)

b: ĐKXĐ: x<>0

\(\frac{x^2-4x+3}{2x}+\frac{x^2+12x+3}{x^2+3}=4\)

=>\(\frac{x^2+3}{2x}-2+\frac{12x}{x^2+3}+1=4\)

=>\(\frac{x^2+3}{2x}+\frac{12x}{x^2+3}=4+2-1=6-1=5\)

=>\(\frac{\left(x^2+3\right)^2+24x^2}{2x\left(x^2+3\right)}=5\)

=>\(\left(x^2+3\right)^2+24x^2-10x\left(x^2+3\right)=0\)

=>\(\left(x^2+3\right)^2-4x\left(x^2+3\right)-6x\left(x^2+3\right)+24x^2=0\)

=>\(\left(x^2+3\right)\left(x^2+3-4x\right)-6x\left(x^2+3-4x\right)=0\)

=>\(\left(x^2-6x+3\right)\left(x^2-4x+3\right)=0\)

TH1: \(x^2-6x+3=0\)

=>\(x^2-6x+9-6=0\)

=>\(\left(x-3\right)^2=6\)

=>\(\left[\begin{array}{l}x-3=\sqrt6\\ x-3=-\sqrt6\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\sqrt6+3\left(nhận\right)\\ x=-\sqrt6+3\left(nhận\right)\end{array}\right.\)

TH2: \(x^2-4x+3=0\)

=>\(x^2-x-3x+3=0\)

=>(x-1)(x-3)=0

=>x=1(nhận) hoặc x=3(nhận)

giải phaj bỏ ngoặc nhức đầu lắm

\(a, x(x+3)-(2x-1)(x+3)=0\)

\(⇔(x+3)(1-x)=0\)

\(⇔\left[\begin{array}{} x+3=0\\ 1-x=0 \end{array}\right.\)

\(⇔\left[\begin{array}{} x=-3\\ x=1 \end{array}\right.\)

Vậy phương trình có tập nghiệm là S={\(-3; 1\)}

\(b, 3x-5(x+2)=3(4-2x)\)

\(⇔3x-5x-10=12-6x\)

\(⇔3x-5x+6x=12+10\)

\(⇔4x=22\)

\(⇔x=\dfrac{22}{4}\)

Vậy pt có 1 nghiệm là \(x=\dfrac{22}{4}\)

\(c, (4x-3)(5x-6)=(4x-3)(2x-3)\)

\(⇔5x-6=2x-3\)

\(⇔5x-2x=-3+6\)

\(⇔3x=3\)

\(⇔x=1\)

Vậy pt có 1 nghiệm là \(x=1\)

Bạn thật tuyệt vời !

em                                                                                                                                                                                                            ko

biết

\(\frac{5}{3}-\left(2x-\frac{2}{4}\right)\ge x-\left(4x-\frac{3}{6}\right)\)

\(\Leftrightarrow\frac{5}{3}-2x+\frac{1}{2}\ge x-4x+\frac{1}{2}\)

\(\Leftrightarrow x\ge-\frac{5}{3}\)

Ý c cx vậy nha ! Chuyển vế rồi thu gọn lại 

a) \(\left\{{}\begin{matrix}2x+3y=5\\4x-5y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x+6y=10\\4x-5y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+3y=5\\11y=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+3\cdot\dfrac{9}{11}=5\\y=\dfrac{9}{11}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+\dfrac{27}{11}=5\\y=\dfrac{9}{11}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x=\dfrac{28}{11}\\y=\dfrac{9}{11}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{14}{11}\\y=\dfrac{9}{11}\end{matrix}\right.\)

Vậy: \(x=\dfrac{14}{11};y=\dfrac{9}{11}\)