c) x² + 9x = 10

x² + 9x - 10 = 0

=> x² - x + 10x - 10 = 0

=> x(x - 1) + 10(x - 1) = 0

=> (x + 10)(x - 1) = 0

=> \(\orbr{\begin{cases}x=-10\\x=1\end{cases}}\)

d) 2x² + 9x = 35

=> 2x² + 9x - 35 = 0

=> 2x² + 14x - 5x - 35 = 0

=> 2x(x + 7) - 5(x + 7) = 0

=> (x + 7)(2x - 5) = 0

=> \(\orbr{\begin{cases}x=-7\\x=\frac{5}{3}\end{cases}}\)

(x² - 2x - 1)² - 5(x² - 2x - 1) - 14 = 0

=> (x² - 2x - 1)² + 2(x² - 2x - 1) - 7(x² - 2x - 1) - 14 = 0

=> (x² - 2x - 1)(x² - 2x + 1) - 7(x² - 2x + 1) = 0

=> (x² - 2x + 1)(x² - 2x - 8) = 0

=> (x - 1)² (x - 4)(x + 2) = 0

=> x = 1 hoặc x = 4 hoặc x = -2

e) (2k² + 5k + 1)² - 12(2k² + 5k + 1) + 32 = 0

=> (2k² + 5x + 1)² - 4(2k² + 5k + 1) - 8(2k² + 5k + 1) + 32 = 0

=> (2k² + 5k + 1)(2k² + 5k - 3) - 8(2k² + 5k - 3) = 0

=> (2k² + 5k - 3)(2k² + 5k - 7) = 0

=> (2k² + 6k - k - 3)(2k² - 2x + 7k - 7) = 0

=> (k + 3)(2k - 1)(k - 1)(2k + 7) = 0

=> k = -3 hoặc k = 1/2 hoặc k = 1 hoặc k = -7/2

1.x² + 6x = 0 < như này nhỉ ? >

⇔ x( x + 6 ) = 0

⇔ x = 0 hoặc x + 6 = 0

⇔ x = 0 hoặc x = -6

2. x² - 25x + 250 = 0

⇔ ( x² - 25x + 625/4 ) + 375/4 = 0

⇔ ( x - 25/2 )² = -375/4 ( vô lí )

=> Phương trình vô nghiệm

3. x² + 9x = 10

⇔ x² + 9x - 10 = 0

⇔ x² - x + 10x - 10 = 0

⇔ x( x - 1 ) + 10( x - 1 ) = 0

⇔ ( x - 1 )( x + 10 ) = 0

⇔ x - 1 = 0 hoặc x + 10 = 0

⇔ x = 1 hoặc x = -10

4. 2x² + 9x = 35

⇔ 2x² + 9x - 35 = 0

⇔ 2x² + 14x - 5x - 35 = 0

⇔ 2x( x + 7 ) - 5( x + 7 ) = 0

⇔ ( x + 7 )( 2x - 5 ) = 0

⇔ x + 7 = 0 hoặc 2x - 5 = 0

⇔ x = -7 hoặc x = 5/2

5. ( x² - 2x - 1 )² - 5( x² - 2x - 1 ) - 14 = 0

Đặt t = x² - 2x - 1

bthuc ⇔ t² - 5t - 14 = 0

          ⇔ t² - 7t + 2t - 14 = 0

          ⇔ t( t - 7 ) + 2( t - 7 ) = 0

          ⇔ ( t - 7 )( t + 2 ) = 0

          ⇔ ( x² - 2x - 1 - 7 )( x² - 2x - 1 + 2 ) = 0

          ⇔ ( x² - 4x + 2x - 8 )( x - 1 )² = 0

          ⇔ ( x - 4 )( x + 2 )( x - 1 )² = 0

          ⇔ x - 4 = 0 hoặc x + 2 = 0 hoặc x - 1 = 0

          ⇔ x = 4 hoặc x = -2 hoặc x = 1

6. ( 2k² + 5k + 1 )² - 12( 2k² + 5k + 1 ) + 32 = 0

Đặt t = 2k² + 5k + 1

bthuc ⇔ t² - 12t + 32 = 0

          ⇔ t² - 8t - 4t + 32 = 0

          ⇔ t( t - 8 ) - 4( t - 8 ) = 0

          ⇔ ( t - 8 )( t - 4 ) = 0

          ⇔ ( 2k² + 5k + 1 - 8 )( 2k² + 5k + 1 - 4 ) = 0

          ⇔ ( 2k² - 2k + 7k - 7 )( 2k² - k + 6k - 3 ) = 0

          ⇔ ( k - 1 )( 2k + 7 )( 2k - 1 )( k + 3 ) = 0

          ⇔ k = 1 hoặc k = -7/2 hoặc k = 1/2 hoặc k = -3

\(\left(4-3x\right)\left(10x-5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}4-3x=0\\10x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=4\\10x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{1}{2}\end{cases}}}\)

\(\left(7-2x\right)\left(4+8x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}7-2x=0\\4+8x=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=7\\8x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}}}\)

rồi thực hiện đến hết ... 

Brainchild bé ngây thơ qus e , ko thực hiện đến hết như thế đc đâu :>

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

\(2x^2-7x+3=4x^2+4x-3\)

\(2x^2-7x+3-4x^2-4x+3=0\)

\(-2x^2-11x+6=0\)

\(2x^2+11x-6=0\)

\(2x^2+12x-x-6=0\)

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

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

\(x+6=0\Leftrightarrow x=-6\)

\(2x-1=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)

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

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

\(x=0\)

\(2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)

Tự lm tiếp nha 

chiu lop 3 ma

a) \(\left(2x-5\right)^2-x^2=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{5}{3}\end{matrix}\right.\)

b) \(\left(x^2-2x+1\right)-9x^2=0\)

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

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

\(\left[{}\begin{matrix}-2x=1\\4x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{4}\end{matrix}\right.\)

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

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

=>x=5 hoặc x=5/3

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

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

=>x=-1/2 hoặc x=1/4

Bài 1:

a)-x^2+4x-5

=-(x²-4x+5)<0 với mọi x

=>-x^2+4x-5<0 với mọi x

b)x^4+3x^2+3

\(=\left(x^2+\frac{3}{2}\right)^2+\frac{3}{4}>0\)với mọi x

=>x^4+3x^2+3>0 với mọi x

c) bn xét từng th ra

Bài 2:

a)9x^2-6x-3=0

=>3(3x²-2x-1)=0

=>3x²-2x-1=0

=>3x²+x-3x-1=0

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

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

b)x^3+9x^2+27x+19=0

=>(x+1)(x²+8x+19) (dùng pp nhẩm nghiệm rồi mò ra)

c)x(x-5)(x+5)-(x+2)(x^2-2x+4)=3

=>x³-25x-x³-8=3

=>-25x-8=3

=>-25x=1

=>x=-11/25

mk sửa 1 tí ở dấu => thứ 2 từ dưới lên là

=>-25x=11

a) 2x (x-5) -(x²-10x +25)=0

\(\Leftrightarrow\)2x(x-5)-(x-5)²=0

\(\Leftrightarrow\)(x-5)(2x-x+5)=0

\(\Leftrightarrow\)(x-5)(x+5)=0

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

\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)

b) x² - 9 +3x(x+3) = 0

\(\Leftrightarrow\)(x² - 9) +3x(x+3) =0

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

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

\(\Leftrightarrow\)(x+3)(4x-3)=0

\(\Leftrightarrow\)\(\left[{}\begin{matrix}x+3=0\\4x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=-3\\4x=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\frac{3}{4}\end{matrix}\right.\)

c) x³ - 16x = 0

\(\Leftrightarrow\)x(x²-16)=0

\(\Leftrightarrow\)x(x-4)(x+4)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)

d) (2x+3)(x-2) - (x² -4x+4) = 0

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

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

\(\Leftrightarrow\)(x-2)(x+5)=0

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

\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)

e) 9x² -(x² -2x +1)=0

\(\Leftrightarrow\)(3x)²-(x-1)²=0

\(\Leftrightarrow\)(3x-x+1)(3x+x-1)=0

\(\Leftrightarrow\)(2x+1)(4x-1)=0

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

\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x=-1\\4x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=\frac{1}{4}\end{matrix}\right.\)

f)x³-4x² -9x +36 = 0

\(\Leftrightarrow\)(x³-9x)-(4x²-36)=0

\(\Leftrightarrow\)x(x²-9)-4(x²-9)=0

\(\Leftrightarrow\)(x-4)(x²-9)=0

\(\Leftrightarrow\)(x-4)(x-3)(x+3)=0

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-3=0\\x+3=0\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=4\\x=3\\x=-3\end{matrix}\right.\)

g) 3x - 6 = (x-1).(x-2)

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

\(\Leftrightarrow\)x-1=3

\(\Leftrightarrow\)x=4

i) (x-2).(x+2) +(2x+1)² =-5x.(x-3) =5 (?? đề sao vậy ??)

k) x² -1 = (x-1).(2x+3)

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

\(\Leftrightarrow\)x+1=2x+3

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

\(\Leftrightarrow\)-x=2

\(\Leftrightarrow\)x=-2

l) (2x-1)² +(x+3).(x-3) -5x(x-2)=6

\(\Leftrightarrow\)4x²-4x+1+x²-9-5x²+10x=6

\(\Leftrightarrow\)6x-8=6

\(\Leftrightarrow\)6x=14

\(\Leftrightarrow\)x=\(\frac{7}{3}\)

a) Ta có: \(x^2-2x-3=0\)

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

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

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

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

Vậy: \(S_1=\left\{3;-1\right\}\)(1)

Ta có: \(\left(x+1\right)\left(x+3\right)=0\)

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

Vậy: \(S_2=\left\{-3;-1\right\}\)(2)

Từ (1) và (2) suy ra \(S_1\ne S_2\)

hay Hai phương trình \(x^2-2x-3=0\) và \(\left(x+1\right)\left(x+3\right)=0\) không tương đương với nhau