ĐKXĐ: x∉{0;2}

\(\frac{x+2}{x}=\frac{2x+3}{2\left(x-2\right)}\)

=>x(2x+3)=2(x-2)(x+2)

=>\(2x^2+3x=2\left(x^2-4\right)=2x^2-8\)

=>3x=-8

=>\(x=-\frac83\) (nhận)

ĐKXĐ: x∉{0;2} x + 2 x = 2 x + 3 2 ( x − 2 ) =>x(2x+3)=2(x-2)(x+2) => 2 x 2 + 3 x = 2 ( x 2 − 4 ) = 2 x 2 − 8 =>3x=-8 => x = − 8 3 (nhận)

Các bước giải chi tiết: 1. Điều kiện xác định (ĐKXĐ): \(x\ne 0\) và \(x\ne 2\) 2. Quy đồng mẫu và khử mẫu: \(\frac{2(x-2)(x+2)}{2x(x-2)}=\frac{x(2x+3)}{2x(x-2)}\)Khử mẫu ta được: \(2(x-2)(x+2)=x(2x+3)\) 3. Khai triển và rút gọn phương trình: \(2(x^{2}-4)=2x^{2}+3x\) \(2x^{2}-8=2x^{2}+3x\) \(3x=-8\) \(x=-\frac{8}{3}\) 4. Kết luận: So sánh với điều kiện xác định (\(x\ne 0\) và \(x\ne 2\)), ta thấy \(x=-\frac{8}{3}\) thỏa mãn.Vậy tập nghiệm của phương trình là: \(S = \left\{ -\frac{8}{3} \right\}\)

a: \(\left(2x-1\right)^2=49\)

=>\(\left[\begin{array}{l}2x-1=7\\ 2x-1=-7\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=8\\ 2x=-6\end{array}\right.\Rightarrow\left[\begin{array}{l}x=4\\ x=-3\end{array}\right.\)

b: (x-2)(3x+5)=(2x-4)(x+1)

=>(x-2)(3x+5)=(x-2)(2x+2)

=>(x-2)(3x+5-2x-2)=0

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

=>\(\left[\begin{array}{l}x-2=0\\ x+3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ x=-3\end{array}\right.\)

c: \(\left(2x+5\right)\left(x-4\right)=\left(x-5\right)\left(4-x\right)\)

=>\(\left(2x+5\right)\left(x-4\right)=\left(-x+5\right)\left(x-4\right)\)

=>\(\left(2x+5\right)\left(x-4\right)-\left(-x+5\right)\left(x-4\right)=0\)

=>(x-4)(2x+5+x-5)=0

=>3x(x-4)=0

=>x(x-4)=0

=>x=0 hoặc x=4

d: \(2\left(9x^2+6x+1\right)=\left(3x+1\right)\left(x-2\right)\)

=>\(2\left(3x+1\right)^2=\left(3x+1\right)\left(x-2\right)\)

=>\(\left(3x+1\right)\left(6x+2-x+2\right)=0\)

=>(3x+1)(5x+4)=0

=>\(\left[\begin{array}{l}3x+1=0\\ 5x+4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac13\\ x=-\frac45\end{array}\right.\)

e: \(\left(2x+7\right)^2=9\left(x+2\right)^2\)

=>\(\left(3x+6\right)^2=\left(2x+7\right)^2\)

=>\(\left(3x+6\right)^2-\left(2x+7\right)^2=0\)

=>(3x+6+2x+7)(3x+6-2x-7)=0

=>(x-1)(5x+13)=0

=>\(\left[\begin{array}{l}x-1=0\\ 5x+13=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\\ x=-\frac{13}{5}\end{array}\right.\)

f: \(27x^2\left(x+3\right)-12\left(x^2+3x\right)=0\)

=>\(27x^2\left(x+3\right)-12x\left(x+3\right)=0\)

=>x(x+3)(27x-12)=0

=>x(x+3)(9x-4)=0

=>\(\left[\begin{array}{l}x=0\\ x+3=0\\ 9x-4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=-3\\ x=\frac49\end{array}\right.\)

g: \(16x^2-8x+1=4\left(x+3\right)\left(4x-1\right)\)

=>\(\left(4x-1\right)^2-\left(4x-1\right)\left(4x+12\right)=0\)

=>(4x-1)(4x-1-4x-12)=0

=>-13(4x-1)=0

=>4x-1=0

=>4x=1

=>x=1/4

h: \(\left(5x^2-2x+10\right)^2=\left(3x^2+10x-8\right)^2\)

=>\(\left(5x^2-2x+10-3x^2-10x+8\right)\left(5x^2-2x+10+3x^2+10x-8\right)=0\)

=>\(\left(2x^2-12x+18\right)\left(8x^2+8x+2\right)=0\)

=>\(2\left(x^2-6x+9\right)\cdot2\cdot\left(4x^2+4x+1\right)=0\)

=>\(\left(x-3\right)^2\cdot\left(2x+1\right)^2=0\)

=>\(\left[\begin{array}{l}x-3=0\\ 2x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=-\frac12\end{array}\right.\)

a: ĐKXĐ: x<>5

\(\frac{4x-3}{x-5}=\frac{29}{3}\)

=>3(4x-3)=29(x-5)

=>29x-145=12x-9

=>17x=-9+145=136

=>x=8(nhận)

b: ĐKXĐ: x<>5/3

\(\frac{2x-1}{5-3x}=2\)

=>2(5-3x)=2x-1

=>10-6x=2x-1

=>-8x=-11

=>\(x=\frac{11}{8}\) (nhận)

c: ĐKXĐ: x<>1

\(\frac{4x-5}{x-1}=2+\frac{x}{x-1}\)

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

=>4x-5=3x-2

=>4x-3x=-2+5

=>x=3(nhận)

d: ĐKXĐ: x∉{-2;5}

\(\frac{7}{x+2}=\frac{3}{x-5}\)

=>7(x-5)=3(x+2)

=>7x-35=3x+6

=>4x=41

=>x=41/4(nhận)

e: ĐKXĐ: x∉{0;-5}

\(\frac{2x+5}{2x}-\frac{x}{x+5}=0\)

=>\(\frac{\left(2x+5\right)\left(x+5\right)-2x^2}{2x\left(x+5\right)}=0\)

=>\(\left(2x+5\right)\left(x+5\right)-2x^2=0\)

=>\(2x^2+10x+5x+25-2x^2=0\)

=>15x=-25

=>\(x=-\frac53\) (nhận)

f: ĐKXĐ: x<>4/11

\(\frac{12x+1}{11x-4}+\frac{10x-4}{9}=\frac{20x+17}{18}\)

=>\(\frac{12x+1}{11x-4}=\frac{20x+17}{18}-\frac{10x-4}{9}=\frac{20x+17-20x+8}{18}=\frac{25}{18}\)

=>25(11x-4)=18(12x+1)

=>275x-100=216x+18

=>59x=118

=>x=2(nhận)

Bài 3:

a: (x+2)(x-3)=6

=>\(x^2-x-6=6\)

=>\(x^2-x-12=0\)

=>(x-4)(x+3)=0

=>x=4 hoặc x=-3

b: (2x+1)(x+2)=2

=>\(2x^2+4x+x+2=2\)

=>\(2x^2+5x=0\)

=>x(2x+5)=0

=>\(\left[\begin{array}{l}x=0\\ 2x+5=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=-\frac52\end{array}\right.\)

c:(x+4)(x+3)=2

=>\(x^2+3x+4x+12-2=0\)

=>\(x^2+7x+10=0\)

=>(x+2)(x+5)=0

=>x=-2 hoặc x=-5

d: (2x+5)(3-x)=14

=>\(6x-2x^2+15-5x=14\)

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

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

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

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

=>x=1 hoặc x=-1/2

Bài 4:

a: \(\left(x+5\right)^2-\left(x+5\right)\left(2x+3\right)=0\)

=>(x+5)(x+5-2x-3)=0

=>(x+5)(2-x)=0

=>\(\left[\begin{array}{l}x+5=0\\ 2-x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-5\\ x=2\end{array}\right.\)

b: \(\left(-3x+5\right)\left(3x-2\right)+\left(3x-2\right)^2=0\)

=>\(\left(3x-2\right)\left(-3x+5+3x-2\right)=0\)

=>3(3x-2)=0

=>3x-2=0

=>3x=2

=>x=2/3

c: \(\left(4-5x\right)\left(3x+\frac12\right)+\left(3x+\frac12\right)^2=0\)

=>\(\left(3x+\frac12\right)\cdot\left(4-5x+3x+\frac12\right)=0\)

=>\(\left(3x+\frac12\right)\left(\frac92-2x\right)=0\)

=>\(\left[\begin{array}{l}3x+\frac12=0\\ \frac92-2x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}3x=-\frac12\\ 2x=\frac92\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac16\\ x=\frac94\end{array}\right.\)

d: \(\left(\frac37+5x\right)\left(-3x-7\right)-\left(-3x-7\right)^2=0\)

=>\(\left(-3x-7\right)\left(5x+\frac37+3x+7\right)=0\)

=>(-3x-7)*(8x+52/7)=0

=>(3x+7)*(8x+52/7)=0

=>\(\left[\begin{array}{l}3x+7=0\\ 8x+\frac{52}{7}=0\end{array}\right.\Rightarrow\left[\begin{array}{l}3x=-7\\ 8x=-\frac{52}{7}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac73\\ x=-\frac{52}{7}:8=-\frac{52}{56}=-\frac{13}{14}\end{array}\right.\)

Xét ΔBEC vuông tại E và ΔBHN vuông tại H có

\(\hat{EBC}\) chung

Do đó: ΔBEC~ΔBHN

=>\(\frac{BE}{BH}=\frac{BC}{BN}\)

=>\(BE\cdot BN=BH\cdot BC\left(1\right)\)

Xét ΔBAC vuông tại A có AH là đường cao

nên \(BH\cdot BC=BA^2=BG^2\left(2\right)\)

Từ (1),(2) suy ra \(BG^2=BE\cdot BN\)

=>\(\frac{BG}{BE}=\frac{BN}{BG}\)

Xét ΔBGN và ΔBEG có

\(\frac{BG}{BE}=\frac{BN}{BG}\)

góc GBN chung

Do đó: ΔBGN~ΔBEG

=>\(\hat{BGN}=\hat{BEG}=90^0\)

=>GB⊥GN