\(\sqrt{7x+7}+\sqrt{7x-6}=t\ge0\)

\(bpt\Leftrightarrow t+t^2< 182\Leftrightarrow-14< t< 13\Leftrightarrow t< 13\Leftrightarrow\sqrt{7x+7}+\sqrt{7x-6}< 13\left(đk:x\ge\dfrac{6}{7}\right)\Leftrightarrow14x+1+2\sqrt{\left(7x+7\right)\left(7x-6\right)}< 169\Leftrightarrow2\sqrt{\left(7x+7\right)\left(7x-6\right)}< 168-14x\Leftrightarrow\left\{{}\begin{matrix}\left(7x+7\right)\left(7x-6\right)\ge0\\168-14x\ge0\\4\left(7x+7\right)\left(7x-6\right)< \left(168-14x\right)^2\end{matrix}\right.\)

\(giảibpt\Rightarrowđáp\) \(số\)

\(x\left(x+6\right)-7x-42=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\x+6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-6\end{cases}}}\)

Bài 1:

a) \(\Rightarrow3x^2+3x-2x^2-4x+x+1=0\)

\(\Rightarrow x^2=-1\left(VLý\right)\Rightarrow S=\varnothing\)

b) \(\Rightarrow\left(x-2020\right)\left(2x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2}\end{matrix}\right.\)

c) \(\Rightarrow\left(x-10\right)\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)

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

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

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

f) \(\Rightarrow\left(5x-4\right)\left(5x+4\right)=0\)

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

Bài 2:

a) \(\Rightarrow3x\left(x^2-4\right)=0\Rightarrow3x\left(x-2\right)\left(x+2\right)=0\)

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

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

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

Ta có: 7x=3y=2z

nên \(\dfrac{x}{\dfrac{1}{7}}=\dfrac{y}{\dfrac{1}{3}}=\dfrac{z}{\dfrac{1}{2}}\)

hay \(\dfrac{7x}{1}=\dfrac{3y}{1}=\dfrac{4z}{2}\)

mà 7x-3y+4z=42

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{7x}{1}=\dfrac{3y}{1}=\dfrac{4z}{2}=\dfrac{7x-3y+4z}{1-1+2}=\dfrac{42}{2}=21\)

Do đó:x=3; y=7; z=10,5

7x + 14 = 42

7x         = 42 - 14

7x         = 28

  x         = 28 : 7

  x         = 4

Vậy x = 4

7x = 42 - 14 

7x = 28

x = 28 : 7

x = 4

7x=42-14

7x=28

x=28:7

x=4

e) \(\left(9x^2-49\right)+\left(3x+7\right)\left(7x+3\right)=0\)

\(\Rightarrow\text{[}\left(3x\right)^2-7^2\text{]}+\left(3x+7\right)\left(7x+3\right)=0\)

\(\Rightarrow\left(3x-7\right)\left(3x+7\right)+\left(3x+7\right)\left(7x+3\right)=0\)

\(\Rightarrow\left(3x+7\right)\text{[}\left(3x-7\right)+\left(7x+3\right)\text{]}=0\)

\(\Rightarrow\left(3x+7\right)\left(3x-7+7x+3\right)=0\)

\(\Rightarrow\left(3x+7\right)\left(10x-4\right)=0\)

=> 2 TH

*3x+7=0               *10x-4=0

=>3x=-7               =>10x=4

=>x=-7/3              =>x=4/10=2/5

vậy x=-7/3 hoặc x=2/5

g) \(\left(x-4\right)^2=\left(2x-1\right)^2\)

\(\Rightarrow\left(x-4\right)^2-\left(2x-1\right)^2=0\)

\(\Rightarrow\left(x-4-2x+1\right)\left(x-4+2x-1\right)=0\)

\(\Rightarrow\left(-x-3\right)\left(3x-5\right)=0\)

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

=> 2 TH

*-(x+3)=0          *3x-5=0

=>-x=-3            =>3x=5  

=x=3                =>x=5/3

h)\(x^2-x^2+x-1=0\)

\(\Rightarrow0+x-1=0\)

\(\Rightarrow x-1=0\)

=>x=0+1

=>x=1

vậy x=1

k, x(x+ 16) - 7x - 42 = 0

=>x^2+16x-7x-42=0

=>x^2+9x-42=0

vì x^2>0

do đó x^2+9x-42>0

nên o có gt nào của x t/m y/cầu đề bài

m)x^2+7x+12=0

=>x^2+3x++4x+12=0

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

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

=>2 TH

=> *x+4=0

=>x=-4

vậy x=-4

*x+3=0

=>x=-3

vậy x=-3

n)x^2-7x+12=0

=>x^2-4x-3x+12=0

=>x(x-4)-3(x-4)=0

=>(x-3).(x-4)=0

=>2 TH

*x-3=0=>x=0+3=>x=3

*x-4=0=>x=0+4=>x=4

vậy x=3 hoặc x=4

a)(3x−3)(5−21x)+(7x+4)(9x−5)=44⇔15x−63x2−15+63x+63x2−35x+36x−20=44⇔79x−35=44⇔79x=79⇒x=1a)(3x−3)(5−21x)+(7x+4)(9x−5)=44⇔15x−63x2−15+63x+63x2−35x+36x−20=44⇔79x−35=44⇔79x=79⇒x=1

b)(x+1)(x+2)(x+5)−x2(x+8)=27⇔x2+2x+x+2(x+5)−x3−8x2=27⇔x2(x+5)+2x(x+5)+x(x+5)+2(x+5)−x3−8x2=27⇔x3+5x2+2x2+10x+x2+5x+2x+10−x3−8x2=27⇔17x+10=27⇔17x=17⇒x=1

\(\left(9x^2-49\right)+\left(3x+7\right)\left(7x+3\right)=0\)

\(\left(9x^2-49\right)+3x\left(7x+3\right)+7\left(7x+3\right)=0\)

\(9x^2-49+21x^2+9x+49x+21=0\)

\(30x^2+58x-28=0\)

\(2\left(15x^2+29x-14\right)=0\)

\(15x^2+29x-14=0\)

.... Tịt :)) 

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

<=> x(x-5) - 4(x-5)=0

<=> (x-5)(x-4)=0

<=>(x-5)=0 hoặc x-4=0

<=> x=5 hoặc x=4

Vậy x={4;5}

b.tương tự

c. x³-5x²+x-5 =0

<=> x²(x-5) + (x-5) = 0

<=> (x-5) (x²+1) = 0

<=> x-5=0 hoặc x²+1=0(loại vì x²=-1)

<=> x=5

vậy x=5

d. bạn kiểm tra lại đề

Tìm x :

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

\(\Leftrightarrow x^2-5x-4x+20=0\)

\(\Leftrightarrow\left(x^2-5x\right)-\left(4x-20\right)=0\)

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

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

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

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

\(\Leftrightarrow x^2+6x-7x-42=0\)

\(\Leftrightarrow\left(x^2+6x\right)-\left(7x+42\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-6\end{matrix}\right.\)

c) \(x^3-5x^2+x-5=0\)

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

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

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

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

d) \(x^4-2x^3+10x^2-20x=0\)

\(\Leftrightarrow\left(x^4-2x^3\right)+\left(10x^2-20x\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x^3+10x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x\left(x^2+10\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\\x^2+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\\x^2=-10\left(vôlí\right)|x^2\ge0\forall x\end{matrix}\right.\)

\(a,\Rightarrow x=150-23=127\\ b,\Rightarrow300:x=42+18=60\\ \Rightarrow x=300:60=5\\ c,\Rightarrow7^{x+3}=7^{23+8}=7^{31}\\ \Rightarrow x+3=31\Rightarrow x=28 \)

a) \(\Rightarrow x=150-23\)

\(\Rightarrow x=127\)

b) \(\Rightarrow300:x=60\)

\(\Rightarrow x=5\)

c) \(\Rightarrow7^{x+3}=7^{31}\)

\(\Rightarrow x+3=31\Rightarrow x=28\)

a.x=127

b x=5

c x=7^28

\(a,PT\Leftrightarrow3x^2+3x-2x^2-4x=-1-x\Leftrightarrow x^2=-1\left(\text{vô nghiệm}\right)\)

Vậy: ...

\(b,PT\Leftrightarrow4x\left(x-2019\right)-\left(x-2019\right)=0\Leftrightarrow\left(x-2019\right)\left(4x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2019\\x=\dfrac{1}{4}\end{matrix}\right.\)

Vậy: ...

\(c,PT\Leftrightarrow\left(x-4-6\right)\left(x-4+6\right)=0\Leftrightarrow\left(x-10\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)

Vậy: ...

\(d,PT\Leftrightarrow\left(x+4\right)^2=0\Leftrightarrow x=-4\)

Vậy: ...

\(e,PT\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)

Vậy: ...

\(f,PT\Leftrightarrow\left(5x-3\right)\left(5x+3\right)=0\Leftrightarrow x=\pm\dfrac{3}{5}\)

Vậy: ...