ĐKXĐ:\(x\ne\pm\dfrac{1}{2}\)

\(\dfrac{1+8x}{4+8x}-\dfrac{4x}{12x-6}+\dfrac{32x^2}{3\left(4-16x^2\right)}=0\)

\(\Leftrightarrow\dfrac{1+8x}{4\left(2x+1\right)}-\dfrac{4x}{6\left(2x-1\right)}+\dfrac{32x^2}{-6\cdot\left(2x-1\right)\left(2x+1\right)}=0\)

\(\Leftrightarrow\dfrac{6\cdot\left(1+8x\right)\left(2x-1\right)}{24\left(2x-1\right)\left(2x+1\right)}-\dfrac{4\cdot4x\left(2x+1\right)}{24\left(2x-1\right)\left(2x+1\right)}-\dfrac{32x^2\cdot4}{24\left(2x-1\right)\left(2x+1\right)}=0\)

\(\Leftrightarrow96x^2-36x-6-36x^2-16x-144x^2=0\)

\(\Leftrightarrow-84x^2-52x-6=0\)

\(\Leftrightarrow\Delta=688\)

\(\Leftrightarrow\left[{}\begin{matrix}x_1=\dfrac{52-\sqrt{688}}{-168}=\dfrac{-13+\sqrt{43}}{42}\\x_2=\dfrac{52+\sqrt{688}}{-168}=\dfrac{-13-\sqrt{43}}{43}\end{matrix}\right.\)

Vậy pt có 2 nghiệm phân biệt............

6x^3 + x + 4 = 11x^2
<=>6x3-11x2+x+4=0
<=>6x3+3x2-14x2-7x+8x+4=0
<=>3x2(2x+1)-7x(2x+1)+4(2x+1)=0
<=>(2x+1)(3x2-7x+4)=0
<=>(2x+1)(3x2-3x-4x+4)=0
<=>(2x+1)(3x-4)(x-1)=0
<=>2x+1=0 hoặc 3x-4=0 hoặc x-1=0
<=>x\(\in\){-1/2;1;4/3}
b)x^6 - 14x^4 + 49x^2 = 36
<=>x6-14x4+49x2-36=0
<=>x6-x4-13x4+13x2+36x2-36=0
<=>x4(x2-1)-13x2(x2-1)+36(x2-1)=0
<=>(x2-1)(x4-13x2+36)=0
<=>(x+1)(x-1)(x4-9x2-4x2+36)=0
<=>(x+1)(x-1)[x2(x2-9)-4(x2-9)]=0
<=>(x-1)(x+1)(x2
-9)(x2-4)=0
<=>(x-1)(x+1)(x+3)(x-3)(x+2)(x-2)=0
<=>x\(\in\){-3;-2;-1;1;2;3}

p/s: kham khảo

6x^3 + x + 4 = 11x^2

<=>6x³-11x²+x+4=0

<=>6x³+3x²-14x²-7x+8x+4=0

<=>3x²(2x+1)-7x(2x+1)+4(2x+1)=0

<=>(2x+1)(3x²-7x+4)=0

<=>(2x+1)(3x²-3x-4x+4)=0

<=>(2x+1)(3x-4)(x-1)=0

<=>2x+1=0 hoặc 3x-4=0 hoặc x-1=0

<=>x\(\in\){-1/2;1;4/3}

b)x^6 - 14x^4 + 49x^2 = 36

<=>x⁶-14x⁴+49x²-36=0

<=>x⁶-x⁴-13x⁴+13x²+36x²-36=0

<=>x⁴(x²-1)-13x²(x²-1)+36(x²-1)=0

<=>(x²-1)(x⁴-13x²+36)=0

<=>(x+1)(x-1)(x⁴-9x²-4x²+36)=0

<=>(x+1)(x-1)[x²(x²-9)-4(x²-9)]=0

<=>(x-1)(x+1)(x²-9)(x²-4)=0

<=>(x-1)(x+1)(x+3)(x-3)(x+2)(x-2)=0

<=>x\(\in\){-3;-2;-1;1;2;3}

phù.mệt

2: ĐKXĐ: x>=0

\(\sqrt{3x}-2\sqrt{12x}+\dfrac{1}{3}\cdot\sqrt{27x}=-4\)

=>\(\sqrt{3x}-2\cdot2\sqrt{3x}+\dfrac{1}{3}\cdot3\sqrt{3x}=-4\)

=>\(\sqrt{3x}-4\sqrt{3x}+\sqrt{3x}=-4\)

=>\(-2\sqrt{3x}=-4\)

=>\(\sqrt{3x}=2\)

=>3x=4

=>\(x=\dfrac{4}{3}\left(nhận\right)\)

3: 

ĐKXĐ: x>=0

\(3\sqrt{2x}+5\sqrt{8x}-20-\sqrt{18}=0\)

=>\(3\sqrt{2x}+5\cdot2\sqrt{2x}-20-3\sqrt{2}=0\)

=>\(13\sqrt{2x}=20+3\sqrt{2}\)

=>\(\sqrt{2x}=\dfrac{20+3\sqrt{2}}{13}\)

=>\(2x=\dfrac{418+120\sqrt{2}}{169}\)

=>\(x=\dfrac{209+60\sqrt{2}}{169}\left(nhận\right)\)

4: ĐKXĐ: x>=-1

\(\sqrt{16x+16}-\sqrt{9x+9}=1\)

=>\(4\sqrt{x+1}-3\sqrt{x+1}=1\)

=>\(\sqrt{x+1}=1\)

=>x+1=1

=>x=0(nhận)

5: ĐKXĐ: x<=1/3

\(\sqrt{4\left(1-3x\right)}+\sqrt{9\left(1-3x\right)}=10\)

=>\(2\sqrt{1-3x}+3\sqrt{1-3x}=10\)

=>\(5\sqrt{1-3x}=10\)

=>\(\sqrt{1-3x}=2\)

=>1-3x=4

=>3x=1-4=-3

=>x=-3/3=-1(nhận)

6: ĐKXĐ: x>=3

\(\dfrac{2}{3}\sqrt{x-3}+\dfrac{1}{6}\sqrt{x-3}-\sqrt{x-3}=-\dfrac{2}{3}\)

=>\(\sqrt{x-3}\cdot\left(\dfrac{2}{3}+\dfrac{1}{6}-1\right)=-\dfrac{2}{3}\)

=>\(\sqrt{x-3}\cdot\dfrac{-1}{6}=-\dfrac{2}{3}\)

=>\(\sqrt{x-3}=\dfrac{2}{3}:\dfrac{1}{6}=\dfrac{2}{3}\cdot6=\dfrac{12}{3}=4\)

=>x-3=16

=>x=19(nhận)

Đáp án cần chọn là: B

Câu 1: 

\(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=3\)

\(\Leftrightarrow\left|x-1\right|+\left|x-2\right|=3\)(1)

Trường hợp 1: x<1

(1) trở thành 1-x+2-x=3

=>3-2x=3

=>x=0(nhận)

Trường hợp 2: 1<=x<2

(1) trở thành x-1+2-x=3

=>1=3(loại)

Trường hợp 3: x>=2

(1) trở thành x-1+x-2=3

=>2x-3=3

=>2x=6

hay x=3(nhận)

=>\(\frac{\left(x+2\right)^2+2}{x+2}+\frac{\left(x+8\right)^2+8}{x+8}\)=\(\frac{\left(x+4\right)+4}{x+4}+\frac{\left(x+6\right)^2+6}{x+6}\)

=>2x+10+\(\frac{2}{x+2}+\frac{8}{x+8}\)=2x+10+\(\frac{4}{x+4}+\frac{6}{x+6}\)

=>-x\(\left(\frac{1}{x+2}-\frac{1}{x+4}-\frac{1}{x+6}+\frac{1}{x+8}\right)\)=0

=>\(\orbr{\begin{cases}x=0\\\frac{1}{x+2}-.....+\frac{1}{x+8}=0\end{cases}}\)

Voi \(\frac{1}{x+2}-....\)=0 ta co

Dat x+5=t

=>\(\frac{1}{t-3}-\frac{1}{t-1}-\frac{1}{t+1}+\frac{1}{t+3}\)=0

=> \(2t\left(\frac{1}{t^2-1}+\frac{1}{t^2-9}\right)=0\)

=>t=0

=>x=-5

Vay phuong trinh co nghiem x=0;-5

toán lớp 8 mà đi giải phương trình hả má

ĐKXĐ:\(x\ne-2;-4;-6;-8\)

\(\frac{x^2+4x+6}{x+2}+\frac{x^2+16+72}{x+8}=\frac{x^2+8x+20}{x+4}+\frac{x^2+12x+42}{x+6}\)

\(\Leftrightarrow\frac{\left(x+2\right)^2+2}{x+2}+\frac{\left(x+8\right)^2+8}{x+8}=\frac{\left(x+4\right)^2+4}{x+4}+\frac{\left(x+6\right)^2+6}{x+6}\)

\(\Leftrightarrow\frac{2}{x+2}+\frac{8}{x+8}=\frac{4}{x+4}+\frac{6}{x+6}\)

\(\frac{10+32}{\left(x+2\right)\left(x+8\right)}=\frac{10x+48}{\left(x+4\right)\left(x+6\right)}\)

\(\Leftrightarrow\frac{5x+16}{\left(x+2\right)\left(x+8\right)}=\frac{5x+24}{\left(x+4\right)\left(x+6\right)}\)

\(\Leftrightarrow\left(5x+16\right)\left(x+4\right)\left(x+6\right)=\left(5x+24\right)\left(x+2\right)\left(x+8\right)\)

\(\Leftrightarrow x^2+5x=0\)(bạn tự biến đổi nhé)

\(\Leftrightarrow x=0;-5\)(tm ĐKXĐ)

Vậy phương trình có nghiệm 0;-5 (mình làm hơi tắt bn thông cảm nha)