Register Now

Login

Lost Password

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

222-9+11+12:2*14+14 = ? ( )

Toán Lớp 8: 1)Tìm x biết a) x(x+5)=0 b) 3x^2-6x=0 c) x(x-6)+10(6-x)=0 2)phân tích thành nhân tử a) x^4-2 x^2y+y^2 b)1-(x^2+2xy+y^2) c)64a^3+125b^3

Tháng Bảy 3, 2022 Toán học 2 Comments 0 views

Toán Lớp 8: 1)Tìm x biết a) x(x+5)=0
b) 3x^2-6x=0
c) x(x-6)+10(6-x)=0
2)phân tích thành nhân tử
a) x^4-2 x^2y+y^2
b)1-(x^2+2xy+y^2)
c)64a^3+125b^3
3) tìm x
a) x^2-81=0
b)4x^3-36=0
c)x^2-10x+25=0

Comments ( 2 )

  1. kimnguenoanh
    kimnguenoanh
    3 Tháng Bảy, 2022 at 8:27 sáng
    Reply
    Bài 1:
    \(a)\,x(x+5)=0\\\Leftrightarrow\left[ \begin{array}{l}x=0\\x+5=0\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=x=0\\x=-5\end{array} \right.\)
    Vậy \(S=\left\{{0;-5}\right\}\)
    \(b)\,3x^2-6x=0\\\Leftrightarrow x(3x-6)=0\\\Leftrightarrow \left[ \begin{array}{l}x=0\\3x-6=0\end{array} \right. \\\Leftrightarrow \left[ \begin{array}{l}x=0\\x=\dfrac{6}{3}=2\end{array} \right.\)
    Vậy \(S=\left\{{0;2}\right\}\)
    \(c)\,x(x-6)+10(6-x)=0\\\Leftrightarrow x(x-6)-10(x-6)=0\\\Leftrightarrow (x-6)(x-10)=0\\\Leftrightarrow\left[ \begin{array}{l}x-6=0\\x-10=0\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=6\\x=10\end{array} \right.\)
    Vậy \(S=\left\{{6;10}\right\}\)
    Bài 2:
    \(a)\,x^4-2 x^2y+y^2\\=(x^2)^2-2.x^2.y+y^2\\=(x^2-y)^2\\b)\,1-(x^2+2xy+y^2)\\=1^2-(x+y)^2\\=[1+(x+y)][1-(x+y)]\\=(1+x+y)(1-x-y)\\c)\,64a^3+125b^3\\=(4a+5a)[(4a)^2-4a.5b+(5b)^2]\\=(4a+5b)(16a^2-20ab+25b^2)\)
    Bài 3:
    \(a)\,x^2-81=0\\\Leftrightarrow x^2=81\\\Leftrightarrow x=\pm9\)
    Vậy \(S=\left\{{9;-9}\right\}\)
    \(b)\,4x^2-36=0\\\Leftrightarrow 4x^2=36\\\Leftrightarrow x^2=9\\\Leftrightarrow x=\pm3\)
    Vậy \(S=\left\{{3;-3}\right\}\)
    \(c)\,x^2-10x+25=0\\\Leftrightarrow x^2-2.5.x+5^2\\\Leftrightarrow (x-5)^2=0\\\Leftrightarrow x-5=0\\\Leftrightarrow x=5\)
    Vậy \(S=\left\{{5}\right\}\)

  2. tuyetnhungthu
    tuyetnhungthu
    3 Tháng Bảy, 2022 at 8:27 sáng
    Reply
    Lời giải:
    1)
    a)
    x(x+5)=0
    $⇔\left[\begin{matrix}x=0\\ x+5=0\end{matrix}\right.$
    $⇔\left[\begin{matrix}x=0\\ x=-5\end{matrix}\right.$
    Vậy x∈{0;-5}
    b)
    3x^(2)-6x=0
    ⇔3x(x-2)=0
    $⇔\left[\begin{matrix}3x=0\\ x-2=0\end{matrix}\right.$
    $⇔\left[\begin{matrix}x=0\\ x=2\end{matrix}\right.$
    vậy x∈{0;2}
    c)
    x(x-6)+10(6-x)=0
    ⇔x(x-6)-10(x-6)=0
    ⇔(x-6)(x-10)=0
    $⇔\left[\begin{matrix}x-6=0\\ x-10=0\end{matrix}\right.$
    $⇔\left[\begin{matrix}x=6\\ x=10\end{matrix}\right.$
    Vậy x∈{6;10}
    2)
    a)
    x^(4)-2x^(2)y+y^2
    =(x^2)^(2)-2.x^(2).y+y^2
    =(x^(2)-y)^2
    b)
    1-(x^(2)+2xy+y^2)
    =1^(2)-(x+y)^2
    =(1-x-y)(1+x+y)
    c)
    64a^(3)+125b^(3)
    =(4a)^(3)+(5b)^3
    =(4a+5b)(16a^(2)-20ab+25b^2)
    3)
    a)
    x^(2)-81
    ⇔x^(2)-9^2=0
    ⇔(x-9)(x+9)=0
    $⇔\left[\begin{matrix}x-9=0\\ x+9=0\end{matrix}\right.$
    $⇔\left[\begin{matrix}x=9\\ x=-9\end{matrix}\right.$
    Vậy x∈{9;-9}
    b)
    4x^(2)-36=0
    ⇔4(x^(2)-9)=0
    ⇔x^(2)-9=0
    ⇔(x-3)(x+3)=0
    $⇔\left[\begin{matrix}x-3=0\\ x+3=0\end{matrix}\right.$
    $⇔\left[\begin{matrix}x=3\\ x=-3\end{matrix}\right.$
    Vậy x∈{3;-3}
    c)
    x^(2)-10x+25=0
    ⇔(x-5)^2=0
    ⇔x-5=0
    ⇔x=5
    Vậy x=5

Leave a reply

222-9+11+12:2*14+14 = ? ( )

Nguyệt Minh

About Nguyệt Minh