Đặt 

Suy ra 

Phương trình đã cho trở thành:

0,05.2u = 3,3 − u ⇔ 0,1u = 3,3 – u ⇔ 1,1u = 3,3 ⇔ u = 3.

Do đó: 

⇔ x – 2010 = 0

⇔ x = 2010.

làm y hệt sách giáo khoa đã dạy

Sách lớp mấy bạn? 

\(a,\left(x-6\right)\left(2x-5\right)\left(3x+9\right)=0\Leftrightarrow\left[{}\begin{matrix}x-6=0\Leftrightarrow x=6\\2x-5=0\Leftrightarrow x=\dfrac{5}{2}\\3x+9=0\Leftrightarrow x=-3\end{matrix}\right.\)

\(b,2x\left(x-3\right)+5\left(x-3\right)=0\Leftrightarrow\left(2x+5\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x-3=0\Leftrightarrow x=3\\2x+5=0\Leftrightarrow x=-\dfrac{5}{2}\end{matrix}\right.\)

\(c,x^2-4-\left(x-2\right)\left(3-2x\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0\Leftrightarrow\left(x-2\right)\left(x+2-3+2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)

\(x=-7\left(2m-5\right)x-2m^2+8\Leftrightarrow x+7\left(2m-5\right)=8-2m^2\Leftrightarrow x\left(14m-34\right)=8-2m^2\)

\(ycđb\Leftrightarrow14m-34\ne0\Leftrightarrow m\ne\dfrac{34}{14}\)\(\Rightarrow x=\dfrac{8-2m^2}{14m-34}\)

\(3.17\Leftrightarrow4x^2-4x+1-2x-1=0\Leftrightarrow4x^2-6x=0\Leftrightarrow x\left(4x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)

3.15:

a, \(\Leftrightarrow\left\{{}\begin{matrix}x-6=0\\2x-5=0\\3x+9=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\x=\dfrac{5}{2}\\x=-\dfrac{9}{3}=-3\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)

3.16

\(\Leftrightarrow\left(2m-5\right).-7-2m^2+8=0\)

\(\Leftrightarrow-14m+35-2m^2+8=0\)

\(\Leftrightarrow-14m-2m^2+43=0\)

\(\Leftrightarrow-2\left(7m+m^2\right)=-43\)

\(\Leftrightarrow m\left(7-m\right)=\dfrac{43}{2}\)

\(\Leftrightarrow\dfrac{m\left(7-m\right)}{1}-\dfrac{43}{2}=0\)

\(\Leftrightarrow\dfrac{14m-2m^2}{2}-\dfrac{43}{2}=0\)

pt vô nghiệm

a) x^2 - 3x + 2 = 0

\(\Delta=b^2-4ac=\left(-3\right)^2-4.1.2=1\)

=> pt có 2 nghiệm pb

\(x_1=\frac{-\left(-3\right)+1}{2}=2\)

\(x_2=\frac{-\left(-3\right)-1}{2}=1\)

a) Dễ thấy phương trình có a + b + c = 0 

nên pt đã cho có hai nghiệm phân biệt x₁ = 1 ; x₂ = c/a = 2

b) \(\hept{\begin{cases}x+3y=3\left(I\right)\\4x-3y=-18\left(II\right)\end{cases}}\)

Lấy (I) + (II) theo vế => 5x = -15 <=> x = -3

Thay x = -3 vào (I) => -3 + 3y = 3 => y = 2

Vậy pt có nghiệm ( x ; y ) = ( -3 ; 2 )

a, (3x+1)(7x+3)=(5x-7)(3x+1)

<=> (3x+1)(7x+3)-(5x-7)(3x+1)=0

<=> (3x+1)(7x+3-5x+7)=0

<=> (3x+1)(2x+10)=0

<=> 2(3x+1)(x+5)=0

=> 3x+1=0 hoặc x+5=0

=> x= -1/3 hoặc x=-5

Vậy...

a) (3x - 2)(4x + 5) = 0

⇔ 3x - 2 = 0 hoặc 4x + 5 = 0

1) 3x - 2 = 0 ⇔ 3x = 2 ⇔ x = 2/3

2) 4x + 5 = 0 ⇔ 4x = -5 ⇔ x = -5/4

Vậy phương trình có tập nghiệm S = {2/3;−5/4}

b) (2,3x - 6,9)(0,1x + 2) = 0

⇔ 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0

1) 2,3x - 6,9 = 0 ⇔ 2,3x = 6,9 ⇔ x = 3

2) 0,1x + 2 = 0 ⇔ 0,1x = -2 ⇔ x = -20.

Vậy phương trình có tập hợp nghiệm S = {3;-20}

c) (4x + 2)(x² +  1) = 0 ⇔ 4x + 2 = 0 hoặc x² +  1 = 0

1) 4x + 2 = 0 ⇔ 4x = -2 ⇔ x = −1/2

2) x² +  1 = 0 ⇔ x² = -1 (vô lí vì x² ≥ 0)

Vậy phương trình có tập hợp nghiệm S = {−1/2}

d) (2x + 7)(x - 5)(5x + 1) = 0

⇔ 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0

1) 2x + 7 = 0 ⇔ 2x = -7 ⇔ x = −7/2

2) x - 5 = 0 ⇔ x = 5

3) 5x + 1 = 0 ⇔ 5x = -1 ⇔ x = −1/5

Vậy phương trình có tập nghiệm là S = {−7/2;5;−1/5}

a, ĐK: \(x\le-1,x\ge3\)

\(pt\Leftrightarrow2\left(x^2-2x-3\right)+\sqrt{x^2-2x-3}-3=0\)

\(\Leftrightarrow\left(2\sqrt{x^2-2x-3}+3\right).\left(\sqrt{x^2-2x-3}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-2x-3}=-\dfrac{3}{2}\left(l\right)\\\sqrt{x^2-2x-3}=1\end{matrix}\right.\)

\(\Leftrightarrow x^2-2x-3=1\)

\(\Leftrightarrow x^2-2x-4=0\)

\(\Leftrightarrow x=1\pm\sqrt{5}\left(tm\right)\)

b, ĐK: \(-2\le x\le2\)

Đặt \(\sqrt{2+x}-2\sqrt{2-x}=t\Rightarrow t^2=10-3x-4\sqrt{4-x^2}\)

Khi đó phương trình tương đương:

\(3t-t^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=0\\t=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2+x}-2\sqrt{2-x}=0\\\sqrt{2+x}-2\sqrt{2-x}=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2+x=8-4x\\2+x=17-4x+12\sqrt{2-x}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\left(tm\right)\\5x-15=12\sqrt{2-x}\left(1\right)\end{matrix}\right.\)

Vì \(-2\le x\le2\Rightarrow5x-15< 0\Rightarrow\left(1\right)\) vô nghiệm

Vậy phương trình đã cho có nghiệm \(x=\dfrac{6}{5}\)

a: \(2x^3-50x=0\)

=>\(2x\left(x^2-25\right)=0\)

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

=>x∈{0;5;-5}

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

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

=>(3x-5)(2x+1)=0

=>\(\left[\begin{array}{l}3x-5=0\\ 2x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac53\\ x=-\frac12\end{array}\right.\)

c: \(9\left(3x-2\right)=x\left(2-3x\right)\)

=>9(3x-2)-x(2-3x)=0

=>9(3x-2)+x(3x-2)=0

=>(3x-2)(x+9)=0

=>\(\left[\begin{array}{l}3x-2=0\\ x+9=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac23\\ x=-9\end{array}\right.\)

d: \(\left(2x-1\right)^2-25=0\)

=>(2x-1-5)(2x-1+5)=0

=>(2x-6)(2x+4)=0

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

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

e: \(25x^2-2=0\)

=>\(25x^2=2\)

=>\(x^2=\frac{2}{25}\)

=>\(\left[\begin{array}{l}x=\frac{\sqrt2}{5}\\ x=-\frac{\sqrt2}{5}\end{array}\right.\)

f: \(x^2-25=6x-9\)

=>\(x^2-6x-16=0\)

=>(x-8)(x+2)=0

=>\(\left[\begin{array}{l}x-8=0\\ x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=8\\ x=-2\end{array}\right.\)

g: 5x(x-3)-2x+6=0

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

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

=>\(\left[\begin{array}{l}x-3=0\\ 5x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=\frac25\end{array}\right.\)

h: 3x(x-7)-2(x-7)=0

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

=>\(\left[\begin{array}{l}x-7=0\\ 3x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=7\\ x=\frac23\end{array}\right.\)

i: \(7x^2-28=0\)

=>\(7x^2=28\)

=>\(x^2=4\)

=>x=2 hoặc x=-2

j: 2x+1+x(2x+1)=0

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

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

k: \(\left(x+2\right)^2-\left(x-2\right)\left(x+2\right)=0\)

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

=>4(x+2)=0

=>x+2=0

=>x=-2

l: \(x^3+5x^2-4x-20=0\)

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

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

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

=>x∈{-5;2;-2}

m: \(x^2-25+2\left(x+5\right)=0\)

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

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

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

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

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

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

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

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

o: \(x^2-6x+8=0\)

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

=>\(\left[\begin{array}{l}x-2=0\\ x-4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ x=4\end{array}\right.\)

p: \(x^2-5x-14=0\)

=>\(x^2-7x+2x-14=0\)

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

=>\(\left[\begin{array}{l}x-7=0\\ x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=7\\ x=-2\end{array}\right.\)

q: \(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)

=>\(x^2-4x+4-x^2+9=6\)

=>-4x+13=6

=>-4x=6-13=-7

=>x=7/4

r: \(\left(2x-1\right)^2-\left(2x-5\right)\left(2x+5\right)=18\)

=>\(4x^2-4x+1-\left(4x^2-25\right)=18\)

=>-4x+26=18