tìm x nha

c, \(3x^2-7x+10=0\)

\(\Leftrightarrow3x^2+3x-10x+10=0\)

\(\Leftrightarrow3x\left(x+1\right)-10\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(3x-10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{10}{3}\end{matrix}\right.\)

d, \(2x\left(x-10\right)-x+10=0\)

\(\Leftrightarrow2x\left(x-10\right)-\left(x-10\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=10\\x=\dfrac{1}{2}\end{matrix}\right.\)

a) x(x - 5) - 4x + 20 = 0

\(\Leftrightarrow\) x(x - 5) - (4x + 20)

\(\Leftrightarrow\) x(x - 5) - 4(x - 5) = 0

\(\Leftrightarrow\) (x - 5)(x - 4)

Khi x - 5 = 0 hoặc x - 4 = 0

 \(\Leftrightarrow\) x = 5           \(\Leftrightarrow\) x = 4

 Vậy S = \(\left\{5;4\right\}\)

b) x(x + 6) - 7x - 42 = 0

 \(\Leftrightarrow\) x(x + 6) - (7x - 42) = 0

 \(\Leftrightarrow\) x(x + 6) - 7(x + 6) = 0

 \(\Leftrightarrow\) (x + 6)(x - 7) = 0

Khi x - 6 = 0 hoặc x - 7 = 0

   \(\Leftrightarrow\) x = 6           \(\Leftrightarrow\) x = 7

 Vậy S = \(\left\{6;7\right\}\)

c) x³ - 5x² - x + 5 = 0

 \(\Leftrightarrow\) (x³ - 5x²) - (x + 5) = 0

 \(\Leftrightarrow\) x² (x - 5) - (x - 5) = 0

 \(\Leftrightarrow\) (x - 5)(x² - 1) = 0

 \(\Leftrightarrow\) (x - 5)(x - 1)(x + 1) = 0

 Khi x - 5 = 0 hoặc x - 1 = 0 hoặc x + 1 = 0

   \(\Leftrightarrow\) x = 5           \(\Leftrightarrow\) x = 1            \(\Leftrightarrow\) x = -1

 Vậy S = \(\left\{5;1;-1\right\}\)

d) 4x² - 25 - (2x - 5)(3x + 7) = 0

\(\Leftrightarrow\) (2x)² - 5² - (2x - 5)(3x + 7) = 0

\(\Leftrightarrow\) (2x - 5)(2x + 5) - (2x - 5)(3x + 7) = 0

\(\Leftrightarrow\) (2x - 5) \([\left(2x+5\right)-\left(3x+7\right)]\) = 0

\(\Leftrightarrow\) (2x - 5) ( 2x + 5 - 3x + 7) = 0

\(\Leftrightarrow\) (2x - 5)( -x + 12) = 0

Khi 2x - 5 = 0 hoặc -x + 12 = 0

  \(\Leftrightarrow\) 2x = 5             \(\Leftrightarrow\)   -x = -12

  \(\Leftrightarrow\) x = \(\dfrac{5}{2}\)              \(\Leftrightarrow\) x = 12

 Vậy S = \(\left\{\dfrac{5}{2};12\right\}\)

e) x³ + 27 + (x + 3)(x - 9) = 0

\(\Leftrightarrow\) x³ - 3³ + (x + 3)(x - 9) = 0

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

\(\Leftrightarrow\) (x - 3) \(\left[\left(x^2-3x+9\right)+\left(x-9\right)\right]\) = 0

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

\(\Leftrightarrow\) (x - 3)(x² - 2x) = 0

\(\Leftrightarrow\) (x - 3)x(x - 2)

 Khi x - 3 = 0 hoặc x = 0 hoặc x - 2 = 0

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

 Vậy S = \(\left\{3;0;2\right\}\)

 Chúc bạn học tốt

a) Ta có: \(x\left(x-5\right)-4x+20=0\)

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

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

b) Ta có: \(x\left(x+6\right)-7x-42=0\)

\(\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\)

\(\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)

c) Ta có: \(x^3-5x^2-x+5=0\)

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

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

d) Ta có: \(4x^2-25-\left(2x-5\right)\left(3x+7\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x+5-3x-7\right)=0\)

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

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

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

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

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

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

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

\(c,\left(x+3\right)\left(x-5\right)=0\)

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

\(d,\left(x+\dfrac{1}{2}\right)\left(4x+4\right)=0\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\5x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

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

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

`a,(x-1)(x+2)=0`

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

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

`b,(x -2)(x -5)=0`

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

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

`c,(x +3)(x -5)=0`

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

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

`d,(x + 1/2)(4x + 4)=0`

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

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

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

`e,(x -4)(5x -10)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\5x-10=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\5x=10\end{matrix}\right.\)

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

`f,(2x -1)(3x +6)=0`

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

\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\3x=-6\end{matrix}\right.\)

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

`g,(2,3x -6,9)(0,1x -2)=0`

\(\Leftrightarrow\left[{}\begin{matrix}2,3x-6,9=0\\0,1x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2,3x=6,9\\0,1x=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=20\end{matrix}\right.\)

a.(x - 1)(x + 2)= 0

<=> x-1=0 hoặc x+2=0

<=> x=1 hoặc x=-2

b.(x -2)(x -5)=0

<=> x-2=0 hoặc x-5=0

<=> x=2 hoặc x=5

c.(x +3)(x -5)=0

<=> x+3=0 hoặc x-5=0

<=> x=-3 hoặc x=5

d.(x + 1/2)(4x + 4)=0

<=> x+1/2=0 hoặc 4x+4=0

<=> x=-1/2 hoặc x=-1

e.(x -4)(5x -10)=0

<=> x-4=0 hoặc 5x-10=0

<=> x=4 hoặc x=2

f.(2x -1)(3x +6)=0

<=> 2x-1=0 hoặc 3x+6=0

<=> x=1/2 hoặc x=-2

g.(2,3x -6,9)(0,1x -2)=0

<=> 2,3x-6,9=0 hoặc 0,1x-2=0

<=> x=3 hoặc x=20

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

\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\4x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{4}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{2}{3};-\dfrac{5}{4}\right\}\)

b) \(\left(2,3x-6,9\right)\left(0,1x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2,3x-6,9=0\\0,1x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-20\end{matrix}\right.\)

c) \(\left(4x+2\right)\left(x^2+1\right)=0\)

Vì \(x^2+1\ge1>0\forall x\)

\(\Rightarrow4x+2=0\)

\(\Leftrightarrow x=-\dfrac{1}{2}\)

Vậy: \(S=\left\{-\dfrac{1}{2}\right\}\)

d) \(\left(2x+7\right)\left(x-5\right)\left(5x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+7=0\\x-5=0\\5x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\\x=-\dfrac{1}{5}\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{7}{2};5;-\dfrac{1}{5}\right\}\)

e) \(\left(x-1\right)\left(2x+7\right)\left(x^2+2\right)=0\)

Vì \(x^2+2\ge2>0\forall x\)

\(\Rightarrow\left(x-1\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\)

f) \(\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\)

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

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

\(\Leftrightarrow\left(3x^2+5x+2\right)\left(-2x+1\right)=0\)

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

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

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

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

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

Vậy: \(S=\left\{-1;-\dfrac{2}{3};\dfrac{1}{2}\right\}\)

Để giải các phương trình này, chúng ta sẽ làm từng bước như sau: 1. 13x(7-x) = 26: Mở ngoặc và rút gọn: 91x - 13x^2 = 26 Chuyển về dạng bậc hai: 13x^2 - 91x + 26 = 0 Giải phương trình bậc hai này để tìm giá trị của x. 2. (4x-18)/3 = 2: Nhân cả hai vế của phương trình với 3 để loại bỏ mẫu số: 4x - 18 = 6 Cộng thêm 18 vào cả hai vế: 4x = 24 Chia cả hai vế cho 4: x = 6 3. 2xx + 98x2022 = 98x2023: Rút gọn các thành phần: 2x^2 + 98x^2022 = 98x^2023 Chia cả hai vế cho 2x^2022: x + 49 = 49x Chuyển các thành phần chứa x về cùng một vế: 49x - x = 49 Rút gọn: 48x = 49 Chia cả hai vế cho 48: x = 49/48 4. (x+1) + (x+3) + (x+5) + ... + (x+101): Đây là một dãy số hình học có công sai d = 2 (do mỗi số tiếp theo cách nhau 2 đơn vị). Số phần tử trong dãy là n = 101/2 + 1 = 51. Áp dụng công thức tổng của dãy số hình học: S = (n/2)(a + l), trong đó a là số đầu tiên, l là số cuối cùng. S = (51/2)(x + (x + 2(51-1))) = (51/2)(x + (x + 100)) = (51/2)(2x + 100) = 51(x + 50) Vậy, kết quả của các phương trình là: 1. x = giá trị tìm được từ phương trình bậc hai. 2. x = 6 3. x = 49/48 4. S = 51(x + 50)

nhầm

a)   \(\left(2x-1\right)^2-25=0\)

⇔ \(\left(2x-1\right)^2-5^2=0\)

⇔  \(\left(2x-1-5\right)\left(2x-1+5\right)=0\)

⇒  \(2x-1-5=0\) hoặc \(2x-1+5=0\)

⇔      \(x=3\)           hoặc  \(x=-2\)

Bài 1: Tìm x

a) (2x-1) ² - 25 = 0

<=> (2x-1)² =  25

<=>  2x-1 = 5  hay 2x-1 =-5

<=>  2x= 6      hay  2x=-4

<=>   x=3     hay    x= -2

Vậy S={3; -2}
b) 3x (x-1) + x - 1 = 0

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

<=> x-1=0  hay  3x+1=0

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

<=> x=1 hay x=\(\dfrac{-1}{3}\)

Vậy S={1;\(\dfrac{-1}{3}\)}

c) 2(x+3) - x ² - 3x = 0

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

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

<=> x+3=0 hay 2-x=0

<=> x=-3  hay  x=2

Vậy S={-3;2}
d) x(x - 2) + 3x - 6 = 0

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

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

<=> x-2=0 hay x+3=0

<=> x=2 hay x=-3

Vậy S={2;-3}
e) 4x ² - 4x +1 = 0

<=> (2x-1)²=0

<=> 2x-1=0

<=> 2x=1

<=> x=\(\dfrac{1}{2}\)

Vậy S={\(\dfrac{1}{2}\)}
f) x +5x²  = 0

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

<=>x=0 hay 1+5x=0

<=> x=0 hay 5x=-1

<=> x=0 hay x= \(\dfrac{-1}{5}\)

Vậy S={0;\(\dfrac{-1}{5}\)}
g) x ²+ 2x -3 = 0

<=> x²-x+3x-3=0

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

<=>  (x-1)(x+3)=0

<=> x-1=0 hay x+3=0

<=> x=1  hay x=-3

Vậy S={1;-3}

b) \(\text{3x (x-1) + x - 1 = 0}\)

\(\Rightarrow3x\left(x-1\right)+\left(x-1\right)=0\)

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

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

c) \(\text{2(x+3) - x ² - 3x = 0}\)

\(\Rightarrow2\left(x+3\right)-x\left(x+3\right)=0\\ \Rightarrow\left(2-x\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}2-x=0\\x+3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

d) \(\text{x(x - 2) + 3x - 6 = 0}\)

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

e)

\(\text{4x ² - 4x +1 = 0}\\ \Rightarrow\left(2x-1\right)^2=0\\ \Rightarrow2x-1=0\\ \Rightarrow x=0,5\)

f) \(\text{x +5x ² = 0}\)

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

viết lại câu g đi bạn

* Trả lời:

\(\left(1\right)\) \(-3\left(1-2x\right)-4\left(1+3x\right)=-5x+5\)

\(\Leftrightarrow-3+6x-4-12x=-5x+5\)

\(\Leftrightarrow6x-12x+5x=3+4+5\)

\(\Leftrightarrow x=12\)

\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)

\(\Leftrightarrow6x-15-6+24x=-3x+7\)

\(\Leftrightarrow6x+24x+3x=15+6+7\)

\(\Leftrightarrow33x=28\)

\(\Leftrightarrow x=\dfrac{28}{33}\)

\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)

\(\Leftrightarrow1-3x-6x+12=-4x-5\)

\(\Leftrightarrow-3x-6x+4x=-1-12-5\)

\(\Leftrightarrow-5x=-18\)

\(\Leftrightarrow x=\dfrac{18}{5}\)

\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)

\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)

\(\Leftrightarrow-x-5x=-7\)

\(\Leftrightarrow-6x=-7\)

\(\Leftrightarrow x=\dfrac{7}{6}\)

\(\left(5\right)\) \(3x\left(2x-1\right)-6x\left(x+2\right)=-3x+4\)

\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)

\(\Leftrightarrow-15x+3x=4\)

\(\Leftrightarrow-12x=4\)

\(\Leftrightarrow x=-\dfrac{1}{3}\)

20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)

Vậy...

1) 16 - 8x = 0 ⇔ 8(2 - x) = 0⇔ 2 - x = 0 ⇔ x = 2

Vậy phương trình có nghiệm là x = 2

một đòn bẫy dài một mét .đặt ở đâu để có thể dùng 3600n có thể nâng tảng đá nặng 120kg?