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)

1/ \(x^3-7x+6=0\)

\(\Leftrightarrow x^3+3x^2-3x^2-9x+2x+6=0\)

\(\Leftrightarrow x^2\left(x+3\right)-3x\left(x+3\right)+2\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+2\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-x-2x+2\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left[x\left(x-1\right)+2\left(x-1\right)\right]=0\)

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

\(\Leftrightarrow\)\(x+3=0\)

hoặc   \(x-1=0\)

hoặc   \(x+2=0\)

\(\Leftrightarrow\)\(x=-3\)

hoặc   \(x=1\)

hoặc   \(x=-2\)

Vậy tập nghiệm của phương trình là : \(S=\left\{-3;1;-2\right\}\)

2/ \(x^3-6x^2-x+30\)

\(\Leftrightarrow x^3+2x^2-8x^2-16x+15x+30=0\)

\(\Leftrightarrow x^2\left(x+2\right)-8x\left(x+2\right)+15\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x^2-8x+15\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x^2-3x-5x+15\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left[x\left(x-3\right)-5\left(x-3\right)\right]=0\)

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

\(\Leftrightarrow\)\(x+2=0\)

hoặc   \(x-3=0\)

hoặc   \(x-5=0\)

\(\Leftrightarrow\)\(x=-2\)

hoặc   \(x=3\)

hoặc   \(x=5\)

Vậy tập nghiệm của phương trình là :\(S=\left\{-2;3;5\right\}\)

3/ \(x^3-9x^2+6x+16=0\)

\(\Leftrightarrow x^3+x^2-10x^2-10x+16x+16=0\)

\(\Leftrightarrow x^2\left(x+1\right)-10x\left(x+1\right)+16\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-10x+16\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-8x-2x+16\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left[x\left(x-8\right)-2\left(x-8\right)\right]=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-8\right)\left(x-2\right)=0\)

\(\Leftrightarrow\)\(x+1=0\)

hoặc  \(x-8=0\)

hoặc  \(x-2=0\)

\(\Leftrightarrow\)\(x=-1\)

hoặc   \(x=8\)

hoặc   \(x=2\)

Vậy tập nghiệm của phương trình là :\(S=\left\{-1;8;2\right\}\)

4/ Đề bài sai ! Sửa lại nhé :

 \(2x^3-x^2+5x+3=0\)

\(\Leftrightarrow2x^3+x^2-2x^2-x+6x+3=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x^2-x+3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\left(tm\right)\\\left(x-\frac{1}{2}\right)^2+\frac{11}{4}=0\left(ktm\right)\end{cases}}\)

Vậy tập nghiệm của phương trình là : \(S=\left\{-\frac{1}{2}\right\}\)

a/. x³ - 9x² +27x - 19 = 0

<=> (x³ - 3.x² .3 + 3.3² .x - 3³) + 8 = 0

<=> (x - 3)³ + 8 = 0

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

<=> (x -1)(x² - 6x+ 9 -2x +6 +4) =0

<=> (x - 1)(x²  - 8x + 19) = 0

<=> x - 1 = 0 => x = 1

Vậy S = {1}

Xem lại đề câu b nha bạn?

c/. x³ + 1 -7x -7 =0 

<=> (x³ + 1) -7(x+1)=0

<=> (x+1)(x²-x+1) -7(x+1)=0

<=> (x+1)(x²-x+1-7)=0

<=> x + 1 = 0 hay x² -x - 6 = 0

<=> x = -1 hay (x² - 3x) + (2x - 6) = 0 

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

<=>                 (x - 3)(x+2) = 0

<=> x = -1 hay x = 3 hay x = -2

Vậy S = {-1;3;-2}

X^{3 -} X²-8X²+8X+19X-19=0

<=>X²(X-1)-8X(X-1)+19(X-1)=0

<=>(X-1)(X²-8X+19)=0

vi X^2-8X+19=(X-4)²+3>3

a, x³-x²- 8x² +8x+19x-19=0
<=> (x-1)(x²- 8x+19)=0
<=> x-1=0 <=> x=1
 

a) ĐKXĐ: \(x\ge0\)

Ta có: \(3\sqrt{18x}-5\sqrt{8x}+4\sqrt{50x}=38\)

\(\Leftrightarrow9\sqrt{2x}-10\sqrt{2x}+20\sqrt{2x}=38\)

\(\Leftrightarrow19\sqrt{2x}=38\)

\(\Leftrightarrow\sqrt{2x}=2\)

\(\Leftrightarrow2x=4\)

hay x=2(thỏa ĐK)

b) ĐKXĐ: \(x\ge0\)

Ta có: \(3\sqrt{12x}-2\sqrt{27x}+4\sqrt{3x}=8\)

\(\Leftrightarrow6\sqrt{3x}-6\sqrt{3x}+4\sqrt{3x}=8\)

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

\(\Leftrightarrow3x=4\)

hay \(x=\dfrac{4}{3}\)

c) ĐKXĐ: \(x\ge5\)

Ta có: \(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\)

\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)

\(\Leftrightarrow2\sqrt{x-5}=4\)

\(\Leftrightarrow\sqrt{x-5}=2\)

\(\Leftrightarrow x-5=4\)

hay x=9

a)

\(3.3\sqrt{2x}-5.2\sqrt{2x}+4.5.\sqrt{2x}=38\\ \Leftrightarrow19\sqrt{2x}=38\\ \Leftrightarrow\sqrt{2x}=2\\ \Leftrightarrow x=2\)

b)

\(3.2.\sqrt{3x}-2.3.\sqrt{3x}+4.\sqrt{3x}=8\\ \Leftrightarrow4\sqrt{3x}=8\\ \Leftrightarrow\sqrt{3x}=2\\\Leftrightarrow x=\dfrac{2^2}{3}=\dfrac{4}{3} \)

c)

\(\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\\ \Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\\ \Leftrightarrow2\sqrt{x-5}=4\\ \Leftrightarrow x-5=4\\ \Leftrightarrow x=9\)

TÍnh theo pp nhẩm nghiệm nha bn

1. =9x²(3x-1)-6x(3x-1)+4(3x-1)

=(3x-1)(9x²-6x+4)

2.=x²(2x+1)-x(2x+1)+3(2x+1)

=(2x+1)(x²-x+3)

3.=4x²(x-3)-x(x-3)+6(x-3)

=(x-3)(4x²-x+6)

Tik nha

Chúc bn học tốt!!!!

#Zon