c: =>5x=40

hay x=8

⇔5x=6²+4

⇔5x=36+4

⇔5x=40

⇔x=8

Đây bạn!!!
c) 9x – 4x = 6¹⁷ : 6¹⁵ + 48 : 12
=> 5x = 6²+4
=> 5x = 36+4
=> 5x = 40
=> x = 8
 

=1/4x12= 3

25%của12

=25%x12

=3

3

Áp dụng dãy tỉ số bằng nhau:

b.

\(\dfrac{x}{2}=\dfrac{y}{-5}=\dfrac{x-y}{2-\left(-5\right)}=\dfrac{-7}{7}=-1\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.\left(-1\right)=-2\\y=-5.\left(-1\right)=5\end{matrix}\right.\)

d.

\(\dfrac{4}{x}=\dfrac{7}{y}\Rightarrow\dfrac{y}{7}=\dfrac{x}{4}=\dfrac{y-x}{7-4}=\dfrac{-12}{3}=-4\)

\(\Rightarrow\left\{{}\begin{matrix}x=4.\left(-4\right)=-16\\y=7.\left(-4\right)=-28\end{matrix}\right.\)

\(M=\left(100-98\right)+\left(96-94\right)+...+\left(16-14\right)+12\\ M=2+2+...+2+12\\ \text{Tổng trên có }\left[\left(100-14\right):2+1\right]:2=22\left(\text{số hạng 2}\right)\\ \Leftrightarrow M=2\cdot22+12=44+12=56\)

b: Ta có: \(5\left(x-1\right)^2-\left(1-x\right)\)

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

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

\(=\left(x-1\right)\left(5x-4\right)\)

a: Ta có: \(5x^2-4xy-x^2y\)

\(=x\left(5x-4y-xy\right)\)

a.

\(\left(x^2-x+1\right)\left(x^2-x+2\right)=12\)

Đặt \(x^2-x+1=y\) ta được:

\(y\left(y+1\right)=12\)

\(\Leftrightarrow y^2+y-12=0\)

\(\Leftrightarrow y^2+4y-3y-12=0\)

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

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

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

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

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

b.

\(3y^3-7y^2-7y+3=0\)

\(\Leftrightarrow3\left(y^3+1\right)-7y\left(y+1\right)=0\)

\(\Leftrightarrow3\left(y+1\right)\left(y^2-y+1\right)-7y\left(y+1\right)=0\)

\(\Leftrightarrow\left(y+1\right)\left(3y^2-3y+3-7y\right)=0\)

\(\Leftrightarrow\left(y+1\right)\left(3y^2-10y+3\right)=0\)

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

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

c.

\(x^2+2y^2=4x+4y-6=0\)

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

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

Do \(\left\{{}\begin{matrix}\left(x-2\right)^2\ge0\\\left(y-1\right)^2\ge0\end{matrix}\right.\) với mọi x;y

\(\Rightarrow\left(x-2\right)^2+2\left(y-1\right)^2\ge0\) ; \(\forall x;y\)

Dấu "=" xảy ra khi và chỉ khi:

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

\(\dfrac{1}{2}\left(x^2+y^2\right)^2-2x^2y^2=\dfrac{1}{2}x^4+x^2y^2+\dfrac{1}{2}y^4-2x^2y^2\\ =\dfrac{1}{2}x^4-x^2y^2+\dfrac{1}{2}y^4=\dfrac{1}{2}\left(x^4-2x^2y^2+y^4\right)\\ =\dfrac{1}{2}\left(x^2-y^2\right)^2\)

\(2\left(x^2+y^2\right)^2-2x^2y^2=2\left(x^4+2x^2y^2+y^4\right)-2x^2y^2\\ =2x^4+4x^2y^2+2y^4-2x^2y^2=2x^4+2x^2y^2+2y^4\\ =2\left(x^4+x^2y^2+y^4\right)\)

Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{4}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{x-5z}{4-5\cdot6}=\dfrac{12}{4-30}=\dfrac{-12}{26}=\dfrac{-6}{13}\)

=>x=-24/13; y=-30/13; z=-36/13

Ta có 

\(\dfrac{x}{4}=\dfrac{y}{5}=\dfrac{z}{6}\)

Áp dụng tính chất dãy tỉ số bằng nhau :

\(\dfrac{x}{4}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{x-5z}{4-5.6}=\dfrac{12}{-26}=-\dfrac{6}{13}\\ \Rightarrow\left\{{}\begin{matrix}x=-\dfrac{6}{13}\times4=-\dfrac{24}{13}\\y=-\dfrac{6}{13}\times5=-\dfrac{30}{13}\\z=-\dfrac{6}{13}\times6=-\dfrac{36}{13}\end{matrix}\right.\)

ta có : `x/4=y/5=z/6` và `x/4=y/5=(5z)/30` và `x-5z=12`

Áp dụng t/c dãy tỉ số bằng nhau ta có :

`x/4=y/5=(5z)/30 =(x-5z)/(4-30)=12/(-26)=-6/13`

`=>x/4 =-6/13=>x=-6/13 . 4=-24/13`

`=>y/5 =-6/13=>y=-6/13 . 5= -30/13`

`=>z/6=-6/13=>z=-6/13 . 6=-36/13`

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