Loại 2: đặt t = ax , t > 0 1. 3x^2 - x 2 3x - x^2 2. ( √2 1 )x - 6.( √2 - 1 )x 1 =0 3. ( √2 1 )x ( 3 2√2 )x 1 - √2 = 0 4. ( √3 ) x 2 3x - 4 = 0 5. 22x^2 - 6x 1 - 17.2x^2 - 3x ...

a)(x-3)^4+(x-5)^4=82 b)(n-6)(n-5)(n-4)(n-3)=5.6.7.8 c)x^4+3x^3-6x^2+6x+4=0 d)x^4-8x^3+21x^2-24x+9=0 e)x^4=2x^2+3x-7/16 f)x^2-4x+3+2/x-6/x^2=0 g)x^5-9/2x^4+13x^2-22x+8=0 h)căn bậc 3(15x-1)+căn bậc 3(x-1)=căn bậc 3(3x+1) j)căn(5x-1)+căn(x+2)=7-x k)căn(3x+1)-căn(6-x)+3x^2-14x-8=0 l)căn(x-2)+căn(4-x)+căn(2x-5)=2x^2-5x i)căn bậc 3(x^2-1)+x=căn(x^3-2) o)(+1)căn(x^2-2x+3)=x^2+1 p)x(x+1)(x-3)+3=căn(4-x)+căn(1+x) ...

#Toán lớp 12

0

TT

Thành Trương

8 tháng 6 2018

#Toán lớp 9

2

PK

Phùng Khánh Linh

9 tháng 6 2018

a) ( x - 3)⁴ + ( x - 5)⁴ = 82

Đặt : x - 4 = a , ta có :

( a + 1)⁴ + ( a - 1)⁴ = 82

⇔ a⁴ + 4a³ + 6a² + 4a + 1 + a⁴ - 4a³ + 6a² - 4a + 1 = 82

⇔ 2a⁴ + 12a² - 80 = 0

⇔ 2( a⁴ + 6a² - 40) = 0

⇔ a⁴ - 4a² + 10a² - 40 = 0

⇔ a²( a² - 4) + 10( a² - 4) = 0

⇔ ( a² - 4)( a² + 10) = 0

Do : a² + 10 > 0

⇒ a² - 4 = 0

⇔ a = + - 2

+) Với : a = 2 , ta có :

x - 4 = 2

⇔ x = 6

+) Với : a = -2 , ta có :

x - 4 = -2

⇔ x = 2

KL.....

b) ( n - 6)( n - 5)( n - 4)( n - 3) = 5.6.7.8

⇔ ( n - 6)( n - 3)( n - 5)( n - 4) = 1680

⇔ ( n² - 9n + 18)( n² - 9n + 20) = 1680

Đặt : n² - 9n + 19 = t , ta có :

( t - 1)( t + 1) = 1680

⇔ t² - 1 = 1680

⇔ t² - 41² = 0

⇔ ( t - 41)( t + 41) = 0

⇔ t = 41 hoặc t = - 41

+) Với : t = 41 , ta có :

n² - 9n + 19 = 41

⇔ n² - 9n - 22 = 0

⇔ n² + 2n - 11n - 22 = 0

⇔ n( n + 2) - 11( n + 2) = 0

⇔ ( n + 2)( n - 11) = 0

⇔ n = - 2 hoặc n = 11

+) Với : t = -41 ( giải tương tự )

TT

Thành Trương

8 tháng 6 2018

@Giáo Viên Hoc24.vn

@Giáo Viên Hoc24h

@Giáo Viên

@giáo viên chuyên

@Akai Haruma

1) (4-3x) (10x-5)=02) (7-2x) (4+8x) = 03) (9-7x) (11-3x) = 04) (7-14x) (x-2) = 05) (2x+1) (x-3) = 06) (8-3x) (-3x+5) = 07) (16-8x) (2-6x) = 08) (x+4) (6x-12) = 09) (11-33x) (x+11) = 010) (x-1/4) (x+5/6) = 011) (7/8-2x) (3x+1/3) = 012) 3x - 2x^2 = 013) 5x + 10x^2 = 014) 4x + 3x^2 = 015) -8x^2 + x =016) 10x^2 - 15x = 017) x^2 -4 =018) 9 - x^2 = 019) x^2 -1 = 020) (x-3) (2x-1) = (2x-1) ( 2x+3)21) (5+4x) (-x+2) = (5+4x) (7+5x)22) (4+x) (x-5) = (3x-8) (x-5) = 0 23) (3x-8) (7-21x) - (9+2x) (7-21x)24)...

LH

Lưu Hạ Vy

15 tháng 3 2020

3

2

๖²⁴ʱTú❄⁀ᶦᵈᵒᶫ

15 tháng 3 2020

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

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

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

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

rồi thực hiện đến hết ...

2U

๖²⁴ʱんuﾘｲú❄✎﹏

15 tháng 3 2020

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

tim x bieta/(3x-5)(2x-1)-(x+2)(6x-1)=0b/ (3x-5)(3x+2)-(3x-1)2=-5c/(3x+2)(x-5)=3(x-1)2-2d/ (x+1)2/3 - (x-2)2/2 = 2x+1/2 (x-3)2/6g/49x2=(3x+2)2h/(3x-4)2-(2x-2)2-3(x-2)(2x-1)=0i/ (x-2)(x2-2x+4)-x(x2+2)=15k/ 6x2-7x-3=0m/(x+5)(x-3)+x2-25=0e/ x3+3x2=4x+12f/...

TV

Tran Vu Kiem Anh

3 tháng 10 2016

2

VA

Vũ Anh Quân

3 tháng 10 2016

de qua

NH

Nguyễn Huyền Trang

6 tháng 8 2018

x.(2.x-1)+1/3-2/3.x=0

GN

gia nhi

13 tháng 4 2020

1) (x+6)(3x-1)+x+6=0

2) (x+4)(5x+9)-x-4=0

3)(1-x)(5x+3)÷(3x-7)(x-1)

4)2x (2x-3)=(3-2x)(2-5x)

5)(2x-7)^2-6(2x-7)(x-3)=0

6)(x-2)(x+1)=x^2-4

7) x^2-5x+6=0

8)2x^3+6x^2=x^2+3x

9)(2x+5)^2=(x+2)^2

1

LN

Linh Nhi

15 tháng 4 2020

1) (x+6)(3x-1)+x+6=0

⇔(x+6)(3x-1)+(x+6)=0

⇔(x+6)(3x-1+1)=0

⇔3x(x+6)=0

2) (x+4)(5x+9)-x-4=0

⇔(x+4)(5x+9)-(x+4)=0

⇔(x+4)(5x+9-1)=0

⇔(x+4)(5x+8)=0

3)(1-x)(5x+3)÷(3x-7)(x-1)

=\(\frac{\left(1-x\right)\left(5x+3\right)}{\left(3x-7\right)\left(x-1\right)}=\frac{\left(1-x\right)\left(5x+3\right)}{\left(7-3x\right)\left(1-x\right)}=\frac{\left(5x+3\right)}{\left(7-3x\right)}\)

Đúng(1)

TD

Tuan Dat

22 tháng 1 2019

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

2/(2x+5)(x-4)=(x-5)(4-x)

3/16x^2-8x+1=4(x+3)(4x-1)

4/27x^2(x+3)-12(×^2+3x)=0

5/2(9x^2+6x+1)=(3x+1)(x-2)

6/(2x-1)^2=49

1

TH

Thierry Henry

22 tháng 1 2019

a. \(\left(2x-1\right)\left(3x+2\right)\left(5-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+2=0\\5-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{-2}{3}\\x=5\end{matrix}\right.\)

\(\Rightarrow S=\left\{\dfrac{1}{2};\dfrac{-2}{3};5\right\}\)

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

\(\Leftrightarrow\left(2x+5\right)\left(x-4\right)+\left(x-5\right)\left(x-4\right)\)

\(\Leftrightarrow3x\left(x-4\right)\)

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

\(\Rightarrow S=\left\{0;4\right\}\)

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

\(\Leftrightarrow\left(4x-1\right)^2-4\left(x+3\right)\left(4x-1\right)=0\)

\(\Leftrightarrow\left(4x-1\right)\left(4x-1-4x-3\right)=0\)

\(\Leftrightarrow-4\left(4x-1\right)=0\Leftrightarrow4x-1=0\Leftrightarrow x=\dfrac{1}{4}\)

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

\(\Leftrightarrow27x^2\left(x+3\right)-12x\left(x+3\right)=0\)

\(\Leftrightarrow x\left(27x-12\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\27x-12=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{9}\\x=-3\end{matrix}\right.\)

\(\Rightarrow S=\left\{0;\dfrac{4}{9};-3\right\}\)

e. \(2\left(9x^2+6x+1\right)=\left(3x+1\right)\left(x-2\right)\)

\(\Leftrightarrow2\left(3x+1\right)^2-\left(3x+1\right)\left(x-2\right)=0\)

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

\(\Leftrightarrow\left(3x+1\right)\left(7x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\7x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=\dfrac{-3}{7}\end{matrix}\right.\)

\(\Rightarrow S=\left\{\dfrac{-1}{3};\dfrac{-3}{7}\right\}\)

g. \(\left(2x-1\right)^2=49\)

\(\Leftrightarrow2x-1=7\Leftrightarrow x=4\)

M

Miss

2 tháng 7 2018

Tìm x

a.(x+2).(x+3)-(x-2).(x+5) = 0

b. (2x+3).(x-4)+(x-5)(x+2) = (3x-5)(x-4)

c. (3x+2)(2x+9)-(x+2)(6x+1) = x+1-(x-6)

d. 3( 2x-1).(3x-1)-(2x-3).(9x-1)=0

1

HT

Hoàng Thị Thanh Huyền

2 tháng 7 2018

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

=> x²+5x+6-x²-3x+10=0

=>2x+16=0

=>2x=-16

=>x=-8

(1) giải pt quy về \(ax^2+bx+c=0\)1) \(x^2=3x\) 2) \(x^2-3x=4\) 3) \(x^4-5x^2+6=0\) 4) \(x^3=9x\)5) \(\left(x+2\right)\left(x-3\right)=x^2-4\) 6) \(\dfrac{x+11}{x^2-1}-\dfrac{x-1}{x+1}=\dfrac{2\left(x+7\right)}{x+1}\)giúp mk vs mk cần...

TT

Thanh Thanh

11 tháng 4 2022

Câu 1. Giải các phườn trình sau: a, 3x+6=0 b, 2x-10=0 c, 3x-7=11 d, 3x-9=0 e, 3x(2-x) =15(x-2) f, (x+5)(x+4)=0 g, x(x+4)=0 h, (2x -4)(x-2)=0 i, (x+1/5)(2x-3)=0 k, x²-4x=0 m, 4x²-1=0 n, x²-6x+9=0 l, (3x-5)²-(x+4)²=0 o, 7x(x+2)-5(x+2)=0 p, 3x(2x-5)-4x+10=0 q, (2-2x)-x²+1=0 r, x(1-3x)=5(1-3x) s, 2x-3/4+x+1/6=3 t, x-3/4-2x+1/3=x/6 u, x+1/13+x+2/12=x+3/11+x+4/10 v, 2x+1/15+2x+2/14=2x+3/13+2x+4/12 Giúp e nha mn. E cảm ơn trc ạ! ...

#Toán lớp 9

5

DT

Đỗ Tuệ Lâm

11 tháng 4 2022

1)

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

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

x= 0

x = 3

2) <=> \(x\left(x-3\right)=4\)

=> \(x=\dfrac{4}{x}+3\)

Đúng(1)

TT

Thư Thư

11 tháng 4 2022

\(2,x^2-3x=4\)

\(\Leftrightarrow x^2-3x-4=0\)

\(\Delta=b^2-4ac=\left(-3\right)^2-4\left(-4\right)=25>0\)

\(\Rightarrow\)Pt có 2 nghiệm pb

\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{3+5}{2}=4\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-3-5}{2}=-1\end{matrix}\right.\)

Vậy \(S=\left\{4;-1\right\}\)

\(3,x^4-5x^2+6=0\)

Đặt \(t=x^2\left(t\ge0\right)\)

Pt trở thành

\(t^2-5t+6=0\)

\(\Delta=b^2-4ac=\left(-5\right)^2-4.6=1>0\)

\(\Rightarrow\)Pt ó 2 nghiệm pb

\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{5+1}{2}=3\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-5-1}{2}-3\end{matrix}\right.\)

\(\Rightarrow t=x^2\Leftrightarrow t=\pm\sqrt{3}\)

Vậy \(S=\left\{\pm\sqrt{3}\right\}\)

Đúng(1)

DP

Dung pham tuan dung

1 tháng 4 2020

5

M

Miinhhoa

1 tháng 4 2020

e, 3x(2-x) =15(x-2)

\(\Leftrightarrow3x\left(2-x\right)-15\left(x-2\right)=0\)

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\-3x-15=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)

Vậy..

f, (x+5)(x+4)=0

\(\Leftrightarrow\left\{{}\begin{matrix}x+5=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\x=-4\end{matrix}\right.\)

Vậy..

g, x(x+4)=0

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

,h, (2x -4)(x-2)=0

\(\Leftrightarrow2\left(x-2\right)\left(x-2\right)=0\)

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

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

i, (x+1/5)(2x-3)=0

\(\Leftrightarrow\left\{{}\begin{matrix}x+\frac{1}{5}=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{-1}{5}\\x=\frac{3}{2}\end{matrix}\right.\)

k, x²-4x=0

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

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

m, 4x²-1=0

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

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

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

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

n, x²-6x+9=0

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

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

<=> x=3

l, (3x-5)²-(x+4)²=0

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

\(\Leftrightarrow\left(2x-9\right)\left(4x-1\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-9=0\\4x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=9\\4x=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{9}{2}\\x=\frac{1}{4}\end{matrix}\right.\)

Vậy ..

o, 7x(x+2)-5(x+2)=0

\(\Leftrightarrow\left(x+2\right)\left(7x-5\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+2=0\\7x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\7x=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x=\frac{5}{7}\end{matrix}\right.\)

Vậy....

p, 3x(2x-5)-4x+10=0

\(\Leftrightarrow3x\left(2x-5\right)-\left(4x-10\right)=0\)

\(\Leftrightarrow3x\left(2x-5\right)-2\left(2x-5\right)=0\)

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{5}{2}\\x=\frac{2}{3}\end{matrix}\right.\)

Vậy...

q, (2-2x)-x²+1=0

\(\Leftrightarrow2\left(1-x\right)-\left(x^2-1^2\right)=0\)

\(\Leftrightarrow2\left(1-x\right)-\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow2\left(1-x\right)+\left(1-x\right)\left(x+1\right)=0\)

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

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

Vậy ....

r, x(1-3x)=5(1-3x)

\(\Leftrightarrow x\left(1-3x\right)-5\left(1-3x\right)=0\)

\(\Leftrightarrow\left(1-3x\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}1-3x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3x=-1\\x=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{3}\\x=5\end{matrix}\right.\)

s, 2x-3/4+x+1/6=3

\(\Leftrightarrow x-\frac{7}{12}=3\Leftrightarrow x=3+\frac{7}{12}=\frac{43}{12}\)

S

✟şin❖

1 tháng 4 2020

r, x(1-3x)=5(1-3x)

➜x(1-3x)-5(1-3x)=0

➜(x-5)(1-3x)=0

➜\(\left[{}\begin{matrix}x-5=0\\1-3x=0\end{matrix}\right.\text{➜}\left[{}\begin{matrix}x=5\\x=\frac{1}{3}\end{matrix}\right.\)

Mk lười lắm mai nha!!!~~~~~~~~~~~~

BC

Bé cu gaming

14 tháng 2 2020

a)(3x+1)^2-x^2+8x-16=0

b)(x+1)(x-1)^2-(x+1)(x-2)^2=0

c)(x+1)(x-3)^2=(3x+1)(2x-5)^2

đ)(x+5)(3x+2)^2=x^2(x+5)

e)x^3-8x^2+21x-18=0

f)x^4+x^2+6x-8=0

11

I

Inequalities

14 tháng 2 2020

\(\left(3x+1\right)^2-x^2+8x-16=0\)

\(\Leftrightarrow\left(3x+1\right)^2-\left(x-4\right)^2=0\)

\(\Leftrightarrow\left(3x+1+x-4\right)\left(3x+1-x+4\right)=0\)

\(\Leftrightarrow\left(4x-3\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}4x-3=0\\2x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=\frac{-5}{2}\end{cases}}\)

NV

Nguyễn Việt Hoàng

14 tháng 2 2020

\(\left(3x+1\right)^2-x^2+8x-16=0\)

\(\Leftrightarrow\left(3x+1\right)^2-\left(x^2-8x+16\right)=0\)

\(\Leftrightarrow\left(3x+1\right)^2-\left(x-4\right)^2=0\)

\(\Leftrightarrow\left(3x+1+x-4\right)\left(3x+1-x+4\right)=0\)

\(\Leftrightarrow\left(4x-3\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}4x-3=0\\2x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=\frac{-5}{2}\end{cases}}\)