\(a,\Leftrightarrow x^2-2x-x^2+5x=6\\ \Leftrightarrow3x=6\\ \Leftrightarrow x=2\)

\(b,\Leftrightarrow x^2-6x+9-x+9=0\\ \Leftrightarrow x^2-7x+18=0\\ \Leftrightarrow\left(x^2-7x+\dfrac{49}{4}\right)+\dfrac{23}{4}=0\\ \Leftrightarrow\left(x-\dfrac{7}{2}\right)^2+\dfrac{23}{4}=0\left(vôlí\right)\)

`a)5x(x-1)-(x+2)(5x-7)=6`

`<=>5x^2-5x-(5x^2-7x+10x-14)=6`

`<=>5x^2-5x-(5x^2+3x-14)=6`

`<=>-8x+14=6`

`<=>8x=8<=>x=1`

Vậy `x=1`

`b)(x+2)^2-(x^2-4)=0`

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

`<=>4x+8=0`

`<=>4x=-8`

`<=>x=-2`

Vậy `x=-2`

a)x=5/2

b)x=-2

a)5x.(x-1)-(x+2).(5x-7)=6

<=> 5x²-5x-(5x²-7x+10x-14)=6

<=>  5x²-5x-5x²+7x-10x+14=6

<=> -8x+14=6

<=> -8x=-8 => x=1

Vậy x=1

b) (x+2)²-(x²-4)=0

<=> (x+2)²-(x²-2²)=0 <=> (x+2)²-(x-2)(x+2)=0

<=> (x+2)[(x+2)-(x-2)]=0

<=> (x+2)(x+2-x+2)=0

<=> (x+2).4=0

=> x+2=0

=> x=-2

Vậy x=-2

Bài 1:

\(a,=6x^2+6x\\ b,=15x^3-10x^2+5x\\ c,=6x^3+12x^2\\ d,=15x^4+20x^3-5x^2\\ e,=2x^2+3x-2x-3=2x^2+x-3\\ f,=3x^2-5x+6x-10=3x^2+x-10\)

Bài 2:

\(a,\Leftrightarrow3x^2+3x-3x^2=6\\ \Leftrightarrow3x=6\Leftrightarrow x=2\\ b,\Leftrightarrow6x^2+3x-6x^2+9x-2x-3=10\\ \Leftrightarrow10x=13\Leftrightarrow x=\dfrac{13}{10}\)

`a)|2x-15|=13`

`**2x-15=13`

`<=>2x=28`

`<=>x=14.`

`**2x-15=-13`

`<=>2x=-2`

`<=>x=-1.`

`b)|7x+3|=66`

`**7x+3=66`

`<=>7x=63`

`<=>x9`

`**7x+3=-66`

`<=>7x=-69`

`<=>x=-69/7`

`c)|5x-2|=0`

`<=>5x-2=0`

`<=>5x=2`

`<=>x=2/5`

\(a,\Leftrightarrow\left[{}\begin{matrix}2x-5=13\\2x-5=-13\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-4\end{matrix}\right.\)

Vậy ...

\(b,\Leftrightarrow\left[{}\begin{matrix}7x+3=66\\7x+3=-66\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-\dfrac{69}{7}\end{matrix}\right.\)

Vậy ...

\(c,\Leftrightarrow5x-2=0\)

\(\Leftrightarrow x=\dfrac{5}{2}\)

Vậy ...

a \(\Rightarrow\left[{}\begin{matrix}2x-5=13\\2x-5=-13\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}2x=18\\2x=-8\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=9\\x=-4\end{matrix}\right.\)

b \(\Rightarrow\left[{}\begin{matrix}7x+3=66\\7x+3=-66\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}7x=63\\7x=-69\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=9\\x=-\dfrac{69}{7}\end{matrix}\right.\)

c \(\Rightarrow5x-2=0\Rightarrow x=\dfrac{2}{5}\)

`x :3*5 = 3/4 :(-5/6)`

`x :15 =3/4*(-6/5)=-9/10`

`x = -9/10 *15 =-27/2`

`x-1*2/2 = 8/x -1.2`

`x- 1*1 = 8/x -2`

`x-8/x = -2+1`

`x-8/x =-1`

`x^2 -8x =-x`

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

`x^2 -7x =0`

`x(x-7) =0`

`=>[(x=0),(x=7):}`

`a, x \div 15=-9/10`

`x=-9/10*14`

`x=-27/2`

`b, (x-1*2)/2=8/(x-1*2)`

\(\left(x-1\cdot2\right)\cdot\left(x-1\cdot2\right)=8\cdot2\)

`(x-1*2)^2=16`

`(x-1*2)^2=(+-4)^2`

\(\Rightarrow\left[{}\begin{matrix}x-1\cdot2=4\\x-1\cdot2=-4\end{matrix}\right.\)

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

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

`=>`\(\left[{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\)

`a) 3-5+(-x+3)=6`

`=>5+(-x+3)=3-6`

`=>5+(-x+3)=-3`

`=>-x+3=-3-5`

`=>-x+3=-8`

`=>-x=-8-3`

`=>-x=-11`

`=>x=11`

__

`b)(-4-x)+(4-15)=-15`

`=>(-4-x)+-11=-15`

`=>-4-x=-15-(-11)`

`=>-4-x=-15+11`

`=>-4-x=-4`

`=>x=-4-(-4)`

`=>x=-4+4`

`=>x=0`

`c)(11+x)-(-11-9)=32`

`=>(11+x)-(-20)=32`

`=>(11+x)+20=32`

`=>11+x=32-20`

`=>11+x=12`

`=>x=12-11`

`=>x=1`

`a)3-5+(-x+3)=6`

`5+(-x+3)=3-6`

`5+(-x+3)=-3`

`-x+3=-3-5`

`-x+3=-8`

`-x=-8-3`

`-x=-11`

`x=11`

`b,(-4-x)+(4-15)=-15`

`(-4-x)+(-11)=-15`

`-4-x=-15-(-11)`

`-4-x=-15+11`

`-4-x=-4`

`x=-4-(-4)`

`x=-4+4`

`x=0`

`c)(11+x)-(-11-9)=32`

`(11+x)-(-20)=32`

`(11+x)+20=32`

`11+x=32-20`

`11+x=12`

`x=12-11`

`x=1`

a ) 3 − 5 + ( − x + 3 ) = 6 ⇒ 5 + ( − x + 3 ) = 3 − 6 ⇒ 5 + ( − x + 3 ) = − 3 ⇒ − x + 3 = − 3 − 5 ⇒ − x + 3 = − 8 ⇒ − x = − 8 − 3 ⇒ − x = − 11 ⇒ x = 11 __ b ) ( − 4 − x ) + ( 4 − 15 ) = − 15 ⇒ ( − 4 − x ) ± 11 = − 15 ⇒ − 4 − x = − 15 − ( − 11 ) ⇒ − 4 − x = − 15 + 11 ⇒ − 4 − x = − 4 ⇒ x = − 4 − ( − 4 ) ⇒ x = − 4 + 4 ⇒ x = 0 c ) ( 11 + x ) − ( − 11 − 9 ) = 32 ⇒ ( 11 + x ) − ( − 20 ) = 32 ⇒ ( 11 + x ) + 20 = 32 ⇒ 11 + x = 32 − 20 ⇒ 11 + x = 12 ⇒ x = 12 − 11 ⇒ x = 1

\(a,2x-5+17=6\\ \Rightarrow2x=-6\\ \Rightarrow x=-3\\ b,\Leftrightarrow10-8+6x=-4\\ \Leftrightarrow6x=-6\Leftrightarrow x=-1\\ d,\Rightarrow-2x=-10\\ \Rightarrow x=5\)

câu c giống câu b nhó

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

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x-2y+3z}{2-2\cdot3+3\cdot5}=\dfrac{33}{11}=3\)

Do đó: x=6; y=9; z=15