Đặt \(x^2-4x=t\)

Phương trình \(\Leftrightarrow\frac{t+12}{t+6}=t+8\Leftrightarrow t+12=\left(t+6\right)\left(t+8\right)\)

\(\Leftrightarrow t+12=t^2+14t+48\Leftrightarrow t^2+13t+36=0\Leftrightarrow\left(t+4\right)\left(t+9\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}t+4=0\\t+9=0\end{cases}\Leftrightarrow\orbr{\begin{cases}t=-4\\t=-9\end{cases}}}\)

Với \(t=-4\Rightarrow x^2-4x+4=0\Rightarrow\left(x-2\right)^2=0\Rightarrow x=2\)

Với \(t=-9\Rightarrow x^2-4x+9=0\)vô nghiệm vì \(\Delta=16-36=-20< 0\)

Vậy phương trình có nghiệm x=2

Mơn bạn nhìu nha

a: ĐKXĐ: x>=-2

\(\sqrt{5x+10}=8-x\)

=>\(\begin{cases}8-x\ge0\\ \left(8-x\right)^2=5x+10\end{cases}\Rightarrow\begin{cases}x\le8\\ x^2-16x+64=5x+10\end{cases}\)

=>\(\begin{cases}-2\le x\le8\\ x^2-21x+54=0\end{cases}\Rightarrow\begin{cases}-2\le x\le8\\ \left(x-3\right)\left(x-18\right)=0\end{cases}\)

=>x=3

b: ĐKXĐ: \(4x^2+x-12\ge0\)

=>\(x^2+\frac14x-3\ge0\)

=>\(x^2+2\cdot x\cdot\frac18+\frac{1}{64}-\frac{193}{64}\ge0\)

=>\(\left(x+\frac18\right)^2\ge\frac{193}{64}\)

=>\(\left[\begin{array}{l}x+\frac18\ge\frac{\sqrt{193}}{8}\\ x+\frac18\le-\frac{\sqrt{193}}{8}\end{array}\right.\Rightarrow\left[\begin{array}{l}x\ge\frac{\sqrt{193}-1}{8}\\ x\le\frac{-\sqrt{193}-1}{8}\end{array}\right.\)

\(\sqrt{4x^2+x-12}=3x-5\)

=>\(\begin{cases}3x-5\ge0\\ \left(3x-5\right)^2=4x^2+x-12\end{cases}\Rightarrow\begin{cases}3x\ge5\\ 9x^2-30x+25-4x^2-x+12=0\end{cases}\)

=>\(\begin{cases}x\ge\frac53\\ 5x^2-31x+37=0\end{cases}\)

\(\Delta=\left(-31\right)^2-4\cdot5\cdot37=221\) >0

=>Phương trình có hai nghiệm phân biệt là

\(\left[\begin{array}{l}x=\frac{31-\sqrt{221}}{2\cdot5}=\frac{31-\sqrt{221}}{10}\left(loại\right)\\ x=\frac{31+\sqrt{221}}{10}\left(nhận\right)\end{array}\right.\)

b) \(\dfrac{x-5}{2017}-1+\dfrac{x-2}{2020}-1=\dfrac{x-6}{2016}-1+\dfrac{x-68}{1954}-1\)

​\(\dfrac{x-2022}{2017}+\dfrac{x-2002}{2020}=\dfrac{x-2022}{2016}+\dfrac{x-2022}{1954}\)​

\(\Leftrightarrow\left(x-2022\right)\left(\dfrac{1}{2017}+\dfrac{1}{2020}-\dfrac{1}{2016}-\dfrac{1}{1954}\right)=0\)

\(\Leftrightarrow x-2022=0\left(\dfrac{1}{2017}+\dfrac{1}{2020}-\dfrac{1}{2016}-\dfrac{1}{1954}\ne0\right)\)

\(\Leftrightarrow x=2022\)

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

\(\Leftrightarrow3x-9+5x-5-x-6=0\)

\(\Leftrightarrow7x=20\)

\(\Rightarrow x=\frac{20}{7}\)

b) \(\left|4x-2\right|=8\Leftrightarrow\orbr{\begin{cases}4x-2=8\\4x-2=-8\end{cases}}\Leftrightarrow\orbr{\begin{cases}4x=10\\4x=-6\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{3}{2}\end{cases}}\)

c) \(-3\left|6x+1\right|=-12\)

\(\Leftrightarrow\left|6x+1\right|=4\Leftrightarrow\orbr{\begin{cases}6x+1=4\\6x+1=-4\end{cases}}\Leftrightarrow\orbr{\begin{cases}6x=3\\6x=-5\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{5}{6}\end{cases}}\)

                                                   Bài giải

a, \(3\left(x-3\right)-5\left(-x+1\right)=x+6\)

\(3x-9+5x-5-x-6=0\)

\(7x-20=0\)

\(7x=20\)

\(x=\frac{20}{7}\)

b, \(\left|4x-2\right|=8\)

\(4x-2=\pm8\)

\(\Rightarrow\orbr{\begin{cases}4x-2=-8\\4x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}4x=-6\\4x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)

Vậy \(x\in\left\{-3\text{ ; }2\right\}\)

c, \(-3\left|6x+1\right|=-12\)

\(\left|6x+1\right|=4\)

\(6x+1=\pm4\)

\(\Rightarrow\orbr{\begin{cases}6x+1=-4\\6x+1=4\end{cases}}\Rightarrow\orbr{\begin{cases}6x=-5\\6x=3\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{5}{6}\\x=\frac{1}{2}\end{cases}}\)

\(\Rightarrow\text{ }x\in\left\{-\frac{5}{6}\text{ ; }\frac{1}{2}\right\}\)

Điều kiện:`x>=2`

Ta có:

`sqrt{x+6}-sqrt{x-2}=(x+6-x+2)/(sqrt{x+6}+sqrt{x-2})`

`=8/(\sqrt{x+6}+sqrt{x-2})`

`pt<=>8/(sqrt{x+6}+sqrt{x-2})(1+sqrt{(x-2)(x+6)})=8`

`<=>(1+sqrt{(x-2)(x+6)})/(sqrt{x+6}+sqrt{x-2})=1`

`<=>1+sqrt{(x-2)(x+6)}=sqrt{x+6}+sqrt{x-2}`

`<=>sqrt{(x-2)(x+6)}-sqrt{x+6}=sqrt{x-2}-1`

`<=>sqrt{x+6}(sqrt{x-2}-1)=sqrt{x-2}-1`

`<=>(sqrt{x-2}-1)(sqrt{x+6}-1)=0`

Vì `x>=2=>x+6>=8=>sqrt{x+6}>=2sqrt2`

`=>sqrt{x+6}-1>=2sqrt2-1>0`

`<=>sqrt{x-2}=1`

`<=>x=3(tm)`

Vậy `S={3}`

a, ĐKXĐ:...

\(\sqrt{5x+10}=8-x\\ \Leftrightarrow5x+10=64-16x+x^2\\ \Leftrightarrow x^2-21x+54=0\)

.....

b, ĐKXĐ:...

\(\sqrt{4x^2+x-12}=3x-5\\ \Leftrightarrow4x^2+x-12=9x^2-30x+25\\ \Leftrightarrow5x^2-31x+37=0\)

.....

Để bình 8 - x lên thì cần phải có ĐK x ≤ 8 nữa nhé! Đi thi ko có đk coi như bỏ :)))

\(4x^2+7xy+3y^2=0\)

=>\(4x^2+4xy+3xy+3y^2=0\)

=>4x(x+y)+3y(x+y)=0

=>(x+y)(4x+3y)=0

TH1: x+y=0

=>y=-x

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

=>\(x^2+4x-\left(-x\right)-6=0\)

=>\(x^2+5x-6=0\)

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

=>x=-6 hoặc x=1

Nếu x=-6 thì y=-x=6

Khi x=1 thì y=-x=-1

TH2: 4x+3y=0

=>4x=-3y

=>x=-3y/4

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

=>\(\left(-\frac{3y}{4}\right)^2-3y-y-6=0\)

=>\(\frac{9}{16}y^2-4y-6=0\)

=>\(9y^2-64y-96=0\) (1)

\(\Delta=\left(-64\right)^2-4\cdot9\cdot\left(-96\right)=7552\)

=>(1) có hai nghiệm phân biệt là:

\(\left[\begin{array}{l}y=\frac{64-\sqrt{7552}}{2\cdot9}=\frac{64-8\sqrt{118}}{18}=\frac{32-4\sqrt{118}}{9}\\ y=\frac{64+\sqrt{7552}}{2\cdot9}=\frac{64+8\sqrt{118}}{18}=\frac{32+4\sqrt{118}}{9}\end{array}\right.\)

Khi \(y=\frac{32-4\sqrt{118}}{9}\) thì \(x=\frac{-3}{4}y=\frac{-3}{4}\cdot\frac{32-4\sqrt{118}}{9}=\frac{-3}{4}\cdot\frac49\cdot\left(8-\sqrt{118}\right)=-\frac13\left(8-\sqrt{118}\right)\)

Khi \(y=\frac{32+4\sqrt{118}}{9}\) thì \(x=\frac{-3}{4}y=\frac{-3}{4}\cdot\frac{32+4\sqrt{118}}{9}=\frac{-3}{4}\cdot\frac49\cdot\left(8+\sqrt{118}\right)=-\frac13\left(8+\sqrt{118}\right)\)

We have \(\frac{1}{2}\left(4x-2\right)=5-\left(6-x\right)\)

\(\Leftrightarrow2x-1=x-1\)

\(\Leftrightarrow3x=0\)

\(\Leftrightarrow x=0\)

So ...

1/2(4x-2)=5-(6-x)

=>2x-1=5-6+x

=>2x-x=5-6+1

=>x=0

Vậy S = {0}

đúng 100% nhé, ko đúng thì ko phải hs lớp 8

b)       \(\frac{x-5}{2017}+\frac{x-2}{2020}=\frac{x-6}{2016}+\frac{x-68}{1954}\)

\(\Leftrightarrow\)\(\frac{x-5}{2017}-1+\frac{x-2}{2020}-1=\frac{x-6}{2016}-1+\frac{x-68}{1954}-1\)

\(\Leftrightarrow\)\(\frac{x-2022}{2017}+\frac{x-2022}{2020}=\frac{x-2022}{2016}+\frac{x-2022}{1954}\)

\(\Leftrightarrow\)\(\left(x-2022\right)\left(\frac{1}{2017}+\frac{1}{2020}-\frac{1}{2016}-\frac{1}{1954}\right)=0\)

\(\Leftrightarrow\)\(x-2022=0\)     (vì 1/2017 + 1/2020 - 1/2016 - 1/1954  \(\ne0\))

\(\Leftrightarrow\)\(x=2022\)

Vậy...

b)       \(\frac{x-5}{2017}+\frac{x-2}{2020}=\frac{x-6}{2016}+\frac{x-68}{1954}\)

\(\Leftrightarrow\)\(\frac{x-5}{2017}-1+\frac{x-2}{2020}-1=\frac{x-6}{2016}-1+\frac{x-68}{1954}-1\)

\(\Leftrightarrow\)\(\frac{x-2022}{2017}+\frac{x-2022}{2020}=\frac{x-2022}{2016}+\frac{x-2022}{1954}\)

\(\Leftrightarrow\)\(\left(x-2022\right)\left(\frac{1}{2017}+\frac{1}{2020}-\frac{1}{2016}-\frac{1}{1954}\right)=0\)

\(\Leftrightarrow\)\(x-2022=0\)     (vì 1/2017 + 1/2020 - 1/2016 - 1/1954  \(\ne0\))

\(\Leftrightarrow\)\(x=2022\)

Vậy,....