D

 Bài `1:`

`b)`

`(x-3).(x-5)=x^{2}-1`

`<=>x^{2}-5x-3x+15=x^{2}-1`

`<=>x^{2}-8x+15-x^{2}+1=0`

`<=>-8x+16=0`

`<=>-8x=-16`

`<=>x=2`

Vậy `S={2}`

`c)`

`x^{3}+x^{3}-x^{2}-1=0`

`<=>2x^{3}-x^{2}-1=0`

`<=>2x^{3}-2x^{2}+x^{2}-1=0`

`<=>2x^{2}.(x-1)+(x-1).(x+1)=0`

`<=>(x-1).(2x^{2}+x+1)=0`

Ta có:

`2x^{2}+x+1`

`=2.(x^{2}+1/2x+ 1/2)`

`=2.[x^{2}+2.x. 1/4+(1/4)^{2}+7/16]`

`=2.[(x+1/4)^{2}+7/16]`

`=2.(x+1/4)^{2}+7/8`

Ta có:

`(x+1/4)^{2}\ge0AAx`

`=>2.(x+1/4)^{2}\ge0AAx`

`=>2(x+1/4)^{2}+7/8>0AAx`

`=>x-1=0`

`<=>x=1`

Vậy `S={1}`

`@Nae`

\(a,\left(2x+1\right)^2-4\left(x+2\right)^2=9\\ \Leftrightarrow4x^2+4x+1-4\left(x^2+4x+4\right)-9=0\\ \Leftrightarrow4x^2-4x^2+4x-16x+1-16-9=0\\ \Leftrightarrow-12x=24\\ \Leftrightarrow x=\dfrac{24}{-12}=-2\\ b,\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\\ \Leftrightarrow x^2+6x+9-\left(x^2+4x-32\right)=1\\ \Leftrightarrow x^2-x^2+6x-4x=1-9-32\\ \Leftrightarrow2x=-40\\ \Leftrightarrow x=-20\\ c,3\left(x+2\right)^2+\left(2x-1\right)^2-7\left(x+3\right)\left(x-3\right)=36\\ \Leftrightarrow3\left(x^2+4x+4\right)+\left(4x^2-4x+1\right)-7\left(x^2-9\right)=36\\ \Leftrightarrow3x^2+12x+12+4x^2-4x+1-7x^2+63=36\\ \Leftrightarrow3x^2+4x^2-7x^2+12x-4x=36-12-1-63\\ \Leftrightarrow8x=-40\\ \Leftrightarrow x=\dfrac{-40}{8}=-5\)

a. \(\dfrac{-3}{x^2-9}+\dfrac{5}{3-x}=\dfrac{2}{x+3}\)

<=> \(\dfrac{-3}{x^2-9}+\dfrac{-5}{x-3}=\dfrac{2}{x+3}\)

<=> \(\dfrac{-3}{x^2-9}+\dfrac{-5\left(x+3\right)}{x^2-9}=\dfrac{2\left(x-3\right)}{x^2-9}\)

<=> \(-3+\left(-5\right)\left(x+3\right)=2\left(x-3\right)\)

<=> -3 + (-5x) + (-15) = 2x - 6

<=> -5x -2x = 15 - 6 + 3

<=> -7x = 12

<=> x = \(\dfrac{-12}{7}\)

Vậy ........

b. \(\left|x+5\right|=2x-1\)

Nếu x \(\ge\) -5 => \(\left|x+5\right|\) = x + 5

Nếu x < -5 => \(\left|x+5\right|\) = -(x + 5)

TH1: Nếu x \(\ge\) -5

<=> x + 5 = 2x - 1

<=> x - 2x = -1 - 5

<=> -x = -6 

<=> x = 6

TH2: Nếu x < -5 

<=> -(x + 5) = 2x - 1

<=> -x - 5 = 2x - 1

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

<=> -4 = 3x

<=> x = \(\dfrac{-4}{3}\)

Vậy .........

c. Bạn tự giải câu này nhé (có thể tách các hạng tử rồi tính)

D

A

22
DD
23
A

Lời giải:
a.

a. $(x-1)(x+2)-(x-3)(x+1)=5x-3$

$\Leftrightarrow (x^2+x-2)-(x^2-2x-3)=5x-3$

$\Leftrightarrow 3x+1=5x-3$

$\Leftrightarrow 4=2x$

$\Leftrightarrow x=2$

b.

$(2x-1)(x+3)-(x-2)(x+3)=3x+1$

$\Leftrightarrow (2x^2+5x-3)-(x^2-4)=3x+1$

$\Leftrightarrow x^2+5x+1=3x+1$

$\Leftrightarrow x^2+2x=0$

$\Leftrightarrow x(x+2)=0$

$\Leftrightarrow x=0$ hoặc $x=-2$

c.

$x^2(x-1)-x(x-1)(x+1)=0$

$\Leftrightarrow x^2(x-1)-(x^2+x)(x-1)=0$

$\Leftrightarrow (x-1)[x^2-(x^2+x)]=0$

$\Leftrightarrow (x-1)(-x)=0$

$\Leftrightarrow x-1=0$ hoặc $-x=0$

$\Leftrightarrow x=1$ hoặc $x=0$

d.

$4x(x-5)-(2x-3)(2x+3)=9$

$\Leftrightarrow 4x^2-20x-(4x^2-9)=9$

$\Leftrightarrow -20x=0$

$\Leftrightarrow x=0$

a: Ta có: \(\left(x-1\right)\left(x+2\right)-\left(x-3\right)\left(x+1\right)=5x-3\)

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

\(\Leftrightarrow-2x+4=0\)

\(\Leftrightarrow2x=4\)

hay x=2

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

\(\Leftrightarrow2x^2+6x-x-3-x^2+4-3x-1=0\)

\(\Leftrightarrow x^2+2x=0\)

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

c: Ta có: \(x^2\left(x-1\right)-x\left(x-1\right)\left(x+1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

d: Ta có: \(4x\left(x-5\right)-\left(2x-3\right)\left(2x+3\right)=9\)

\(\Leftrightarrow4x^2-20x-4x^2+9=9\)

hay x=0

`a,x(x-1)-(x+2)^2=1`

`<=>x^2-x-x^2-4x-4=1`

`<=>-5x=5`

`<=>x=-1`

`b,(x+5)(x-3)-(x-2)^2=-1`

`<=>x^2+2x-15-x^2+4x-4+1=0`

`<=>6x-18=0`

`<=>x-3=0`

`<=>x=3`

`c,x(2x-4)-(x-2)(2x+3)=0`

`<=>2x(x-2)-(x-2)(2x+3)=0`

`<=>(x-2)(2x-2x-3)=0`

`<=>-3(x-2)=0`

`<=>x-2=0`

`<=>x=2`

`d,x(3x+2)+(x+1)^2-(2x-5)(2x+5)=-12`

`<=>3x^2+2x+x^2+2x+1-4x^2+25=-12`

`<=>4x+26=-12`

`<=>4x=-38`

`<=>x=-19/2`

a) Ta có: \(\dfrac{2x-1}{5}-\dfrac{x-2}{3}=\dfrac{x+7}{15}\)

\(\Leftrightarrow\dfrac{3\left(2x-1\right)}{15}-\dfrac{5\left(x-2\right)}{15}=\dfrac{x+7}{15}\)

\(\Leftrightarrow6x-3-5x+10-x-7=0\)

\(\Leftrightarrow0x=0\)(luôn đúng)

Vậy: S={x|\(x\in R\)}

Ai giúp vs

1

C = ( 2a - 2 ) (2a + 2 ) - a ( 3 + 4a ) + 3a + 1

C = 4a² - 4 - 3a - 4a² + 3a + 1

C = -3 ko phụ thuộc của x

2. ( x + 3 ) ( x - 1 ) - x ( x - 5 ) = 11

( x² + 3x - x - 3 ) - x² + 5x = 11

7x = 14

x = 2

\(C=\left(2a-2\right)\left(2a+3\right)-a\left(3+4a\right)+3a+1\)

\(\Leftrightarrow C=2a\left(2a-2\right)+3\left(2a-2\right)-3a-4a^2+3a+1\)

\(\Leftrightarrow C=4a^2-4a+6a-6-3a-4a^2+3a+1\)

\(\Leftrightarrow C=\left(4a^2-4a^2\right)+\left(3a-3a\right)+\left(6a-4a\right)+\left(1-6\right)\)

\(\Leftrightarrow C=0+0+2a-5\)

\(\Leftrightarrow C=2a-5\)

Vậy giá trị của C phụ thuộc vào giá trị của biến

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