b) |x + 1| + |x + 2| + |x + 3| = 4x

Có: \(\left\{{}\begin{matrix}\left|x+1\right|>0\\\left|x+2\right|>0\\\left|x+3\right|>0\end{matrix}\right.\) \(\forall x\)

Do đó, \(4x>0=>x>0\).

Lúc này ta có: \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)=4x\)

=> \(3x+6=4x\)

=> \(4x-3x=6\)

=> \(1x=6\)

=> \(x=6:1\)

=> \(x=6\)

Vậy \(x=6\).

Chúc bạn học tốt!

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

Đặt 2y=z

\(A_{\left(x,z\right)}=x^2+z^2+1-2xz+2x-2z\)

\(A_{\left(x,z\right)}=\left(x^2-xz\right)+\left(z^2-xz\right)+\left(x-z\right)+\left(x-z+1\right)\)

\(A_{\left(x,z\right)}=\left[x\left(x-z\right)+z\left(z-x\right)+\left(x-z\right)\right]+\left(x-z+1\right)\)

\(A_{\left(x,z\right)}=\left[\left(x-z\right)\left(x-z+1\right)\right]+\left(x-z+1\right)\)

\(A_{\left(x,z\right)}=\left(x-z+1\right)\left(x-z+1\right)=\left(x-z+1\right)^2\)

Vậy nghiệm đã thức là: \(x-z+1=0\Leftrightarrow x-2y+1=0\)

p/s: lớp 8 không dài dòng thế này%

\(B=x^2+4y+4y^2+8x+42=\left(x^2+8x+16\right)+\left(4y^2+4y+1\right)+25=\left(x+4\right)^2+\left(2y+1\right)^2+25\ge25\)

Dấu = xảy ra khi x = -4; y = -1/2

\(B=x^2+4y+4y^2+8x+42\)
\(B=x^2+8x+16+4y^2+4y+1+25\)
\(B=\left(x+4\right)^2\left(2y+1\right)^2+25\)
GTNN của B là 25
xảy ra khi (x+4)²=0 hoặc (2y+1)²=0
                 x+4=0     hoặc 2y+1=0
                 x=-4        hoặc 2y=-1
                 x= -4       hoặc   y=-1/2

\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\)

\(\Rightarrow\dfrac{4\left(3x-2y\right)}{16}=\dfrac{3\left(2z-4x\right)}{9}=\dfrac{2\left(4y-3z\right)}{4}\)

\(\Rightarrow\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}=\dfrac{12x-8y+6z-12x+8y-6z}{16+9+4}=\dfrac{0}{16+9+4}=0\)\(\Rightarrow\left\{{}\begin{matrix}3x=2y\\2z=4x\\4y=3z\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}\\\dfrac{x}{2}=\dfrac{z}{4}\\\dfrac{y}{3}=\dfrac{z}{4}\end{matrix}\right.\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{3}{9}=\dfrac{1}{3}\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.\dfrac{1}{3}=\dfrac{2}{3}\\y=3.\dfrac{1}{3}=1\\z=4.\dfrac{1}{3}=\dfrac{4}{3}\end{matrix}\right.\)

\(C=x^2+2x+1\dfrac{1}{2}\\ \Rightarrow C=\left(x^2+2x+1\right)+\dfrac{1}{2}\\ \Rightarrow C=\left(x+1\right)^2+\dfrac{1}{2}\ge\dfrac{1}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow x=-1\)

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

 \(C=x^2+2x+1\dfrac{1}{2}.\\ C=x^2+2x+1+\dfrac{1}{2}.\\ C=\left(x+1\right)^2+\dfrac{1}{2}.\)

Ta có: \(\left(x+1\right)^2\ge0\forall x\in R.\\ \dfrac{1}{2}>0. \)

\(\Rightarrow\left(x+1\right)^2+\dfrac{1}{2}\ge\dfrac{1}{2}.\)

Dấu "=" xảy ra khi \(x+1=0.\Leftrightarrow x=-1.\)

Vậy GTNN của biểu thức C là \(\dfrac{1}{2}\) khi x = -1.

1) ta có: \(x:3=y.15\Rightarrow x\cdot\frac{1}{3}=y.15\Rightarrow\frac{x}{15}=\frac{y}{\frac{1}{3}}\)

ADTCDTSBN

...

2) bn ghi thiếu đề r

3) ta có: \(3x=7y\Rightarrow\frac{x}{7}=\frac{y}{3}=k\Rightarrow\hept{\begin{cases}x=7k\\y=3k\end{cases}}\)

mà xy = 189 => 7k.3k = 189

                          21 k² = 189

                                 k² = 9 = 3² = (-3)² => k = 3 hoặc k  = - 3

TH1: k = 3

x = 7.3 => x  = 21

y = 3.3 => y = 9

...

4) ta có: \(4x=5y\Rightarrow\frac{x}{5}=\frac{y}{4}\Rightarrow\frac{x^2}{25}=\frac{y^2}{16}\)

ADTCDTSBN

...

x=5

y=\(\frac{1}{4}\)

x=5

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