Câu 11:

Ta có: \(\left(2x-3\right)\left(3x+2\right)-\left(2x-3\right)^2=-18\)

\(\Leftrightarrow6x^2+4x-9x-6-4x^2+12x-9=-18\)

\(\Leftrightarrow2x^2+7x+3=0\)

\(\Leftrightarrow\left(x+3\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.\)

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b: Xét ΔABD và ΔBAC có

BA chung

BD=AC

AD=BC

Do đó: ΔABD=ΔBAC

c: ta có: EA+EC=AC

EB+ED=BD

mà AC=BD

và EA=EB

nên EC=ED

\(\left(x-5\right)\left(x+5\right)-\left(x+3\right)^2+3\left(x-2\right)^2=\left(x+1\right)^2-\left(x-4\right)\left(x+4\right)+3x^2\)\(\Leftrightarrow x^2-25-\left(x^2+6x+9\right)+3\left(x^2-4x+4\right)=x^2+2x+1-\left(x^2-4^2\right)+3x^2\)\(\Leftrightarrow x^2-25-x^2-6x-9+3x^2-12x+12=x^2+2x+1-x^2+16+3x^2\)

\(\Leftrightarrow-20x=39\)

\(\Leftrightarrow x=\frac{-39}{20}\)

Vậy \(x=\frac{-39}{20}\)

a: Sửa đề: \(D=\frac{x^2}{x^2+3}+\frac{2x^2}{x^2-3}+\frac{3x^4+9}{9-x^4}\)

\(=\frac{x^2\left(x^2-3\right)+2x^2\left(x^2+3\right)-3x^4-9}{\left(x^2+3\right)\left(x^2-3\right)}\)

\(=\frac{x^4-3x^2+2x^4+6x^2-3x^4-9}{\left(x^2+3\right)\left(x^2-3\right)}=\frac{3x^2-9}{\left(x^2+3\right)\left(x^2-3\right)}\)

\(=\frac{3\left(x^2-3\right)}{\left(x^2+3\right)\left(x^2-3\right)}=\frac{3}{x^2+3}\)

b: \(D=\frac13\)

=>\(\frac{3}{x^2+3}=\frac13=\frac39\)

=>\(x^2+3=9\)

=>\(x^2=6\)

=>\(\left[\begin{array}{l}x=\sqrt6\left(nhận\right)\\ x=-\sqrt6\left(nhận\right)\end{array}\right.\)

c: \(x^2+3\ge3\forall x\) thỏa mãn ĐKXĐ

=>\(\frac{3}{x^2+3}\le\frac33=1\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi x=0

Câu I:

1: Ta có: 4x-3=2x+7

nên 2x=10

hay x=5

2: Ta có: \(\left|x-2\right|=4-2x\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=4-2x\left(x\ge2\right)\\x-2=2x-4\left(x< 2\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=6\\-x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=2\left(loại\right)\end{matrix}\right.\)

3: ta có: \(\left(x-1\right)\left(x-2\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-1\end{matrix}\right.\)

Nhóm 1:(45+15):2=30

lớp 8A có 30 Hs trồng hoa

900

900

Ta có:   4x² + 12x + 9 = 0

   \(\Leftrightarrow\)( 2x + 3)² = 0

    \(\Leftrightarrow\)2x + 3 = 0

    \(\Leftrightarrow\)x = \(\frac{-3}{2}\)

Vậy .........

Đúg r ta... S mk k ngĩ ra nhỉ

Cảm ơn bạn

11)11) 3x(x-5)²-(x+2)³+2(x-1)³-(2x+1)(4x²-2x+1)=3x(x²-10x+25)-(x³+6x²+12x+8)+2(x³-3x²+3x-1)-(8x³+1)=3x³-30x²+75x-x³-6x²-12x-8+2x³-6x²+6x-2-8x³-1=-4x³-42x²+63x-11

nhìn khó thế