a) \(6x+15\times8=12\times\left(19-x\right)\)

\(6x+120=228-12x\)

\(6x+120-228+12x=0\)

\(18x-108=0\)

\(18x=108\)

\(x=6\)

b) \(160-\left(35\div x+3\right)\times15=15\)

\(160-\left(35\div x+3\right)=1\)

\(35\div x+3=159\)

\(35\div x=156\)

\(x=\dfrac{35}{156}\)

c) \(2x-\left(1309\div11-19\right)-2=0\)

\(2x-1309\div11-19=2\)

\(2x-119-19=2\)

\(2x-119=21\)

\(2x=140\)

\(x=70\)

d) \(\left(x-7\right)\times\left(2x-16\right)=0\)

\(x-7=0;2x-16=0\)

\(x=7;2x=16\)

\(x=7;x=8\)

a) $6x+15\times 8=12\times \left(19-x\right)$

$6x+120=228-12x$

$6x+120-228+12x=0$

$18x-108=0$

$18x=108$

$x=6$

b) $160-\left(35÷x+3\right)\times 15=15$

$160-\left(35÷x+3\right)=1$

$35÷x+3=159$

$35÷x=156$

�=35156x=15635​
 

x = 5 nha ^.^

`@` `\text {Ans}`

`\downarrow`

`a,`

`x^2 + 2x + 1 = 9`

`=> x^2 + 2x + 1 - 9 = 0`

`=> x^2 + 2x - 8 = 0`

`=> x^2 + 4x - 2x - 8 = 0`

`=> (x^2 + 4x) - (2x + 8) = 0`

`=> x(x + 4) - 2(x + 4) = 0`

`=> (x-2)(x+4) = 0`

`=>`\(\left[{}\begin{matrix}x-2=0\\x+4=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)

Vậy, `x \in {2; 4}`

`b,`

`x^2 - 1 = 15`

`=> x^2 = 15 + 1`

`=> x^2 = 16`

`=> x^2 = (+-4)^2`

`=> x = +-4`

Vậy, `x \in {4; -4}`

`c)`

`19 - 2x^2 = 1`

`=> 2x^2 = 19 - 1`

`=> 2x^2 = 18`

`=> x^2 = 18 \div 2`

`=> x^2 = 9`

`=> x^2 = (+-3)^2`

`=> x = +-3`

Vậy, `x \in {3; -3}.`

a) Ta có: \(x\left(x-1\right)-x^2+2x=5\)

\(\Leftrightarrow x^2-x-x^2+2x=5\)

hay x=5

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

\(\Leftrightarrow2x\left(x-1\right)-\left(x-1\right)^2=0\)

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

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

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

c) Ta có: \(\left(x+3\right)\cdot\left(x^2-3x+9\right)-x\left(x-2\right)^2=19\)

\(\Leftrightarrow x^3+27-x\left(x^2-4x+4\right)-19=0\)

\(\Leftrightarrow x^3+8-x^3+4x^2-4x=0\)

\(\Leftrightarrow4x^2-4x+8=0\)(Vô lý)

Câu 1 : 65.(-19)+19.(-24)= (-65).19+19.(-24)=19.[(-65)+(-24)]=19.(-89)=-1691                        
Câu 2:a) 7x.(2+x)-7x(x+3)=14                
            =>7x.(2+x-x-3)=14     
            =>7x.(-1)=14     
            =>(-7).x=14     
            =>x=14:(-7)    
            =>x=-2   
Vậy x=-2

x^2 -6x +10 = x^2 -2.x.3 +3^2 +1 = (x-3)^2 +1 
Ma (x-3)^2 >=0 <=> (x-3)^2 +1 >=1>0 (voi moi x) 
b) 4x - x^2 -5 = -(x^2 -4x +5) =-[(x^2 -4x +4)+1] = -[(x-2)^2 +1] 
Ma (x+2)^2 >=0 <=> (x-2)^2 +1 >=1 <=> -[(x-2)^2 +1] <=-1 => -[(x-2)^2 +1] <0 
2) a) P= x^2 -2x +5 = x^2 -2x +1 +4 = (x-1)^2 +4 
Ta co: (x-1)^2 >=0 <=> (x-1)^2 +4 >=4 
Vay gia tri nho nhat P=4 khi x=1 
b) Q= 2x^2 -6x = 2(x^2 -3x) = 2(x^2 - 2.x.3/2 + 9/4 -9/4)= 2[(x-3/2)^2 -9/4] 
Ta co: (x-3/2)^2 >=0 <=>(x-3/2)^2 -9/4 >= -9/4 <=> 2[(x-3/2)^2 -9/4] >= -9/2 
Vay gia tri nho nhat Q= -9/2 khi x= 3/2 
c) M= x^2 +y^2 -x +6y +10 = (x^2 -2.x.1/2 + 1/4) +(y^2 +2.y.3+9)+3/4 
= ( x-1/2)^2 + (y+3)^2 +3/4 
M>= 3/4 
Vay GTNN cua M = 3/4 khi x=1/2 va y=-3 
3)a) A= 4x - x^2 +3 = -(x^2 -4x -3) = -( x^2 -4x+4 -7) =-[(x-2)^2 -7] 
Ta co: (x-2)^2>=0 <=> (x-2)^2 -7 >=-7 <=> -[(x-2)^2 -7] <=7 
Vay GTLN A=7 khi x=2 
b) B= x-x^2 = -(x^2 -2.x.1/2+1/4-1/4) = -[(x-1/2)^2 -1/4] 
GTLN B= 1/4 khi x=1/2 
c) N= 2x - 2x^2 -5 =-2( x^2 -x+5/2) = -2(x^2 - 2.x.1/2 +1/4 +9/4) 
= -2[(x-1/2)^2 +9/4] 
GTLN N= -9/2 khi x=1/2

a) \(\Rightarrow x^2+8x+16-x^2+1=19\)

\(\Rightarrow8x=2\Rightarrow x=\dfrac{1}{4}\)

b) \(\Rightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)

\(\Rightarrow2x=-255\Rightarrow x=-\dfrac{255}{2}\)

a: Ta có: \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=19\)

\(\Leftrightarrow x^2+8x+16-x^2+1=19\)

\(\Leftrightarrow x=\dfrac{1}{4}\)

b: Ta có: \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)

\(\Leftrightarrow2x=-255\)

hay \(x=-\dfrac{255}{2}\)