bai nay bn lay trong sach j z

76

a,x^2-25-(x+5)=0

<=>(x^2-5^2)-(x+5)=0

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

<=>(x+5)(x-5+1)=0

<=>(x+5)(x-6)=0

<=>*x+5=0=>x=-5

*x-6=0=>x=6

x=2,5:(3/3,6)=...

Xét tam giác PMO và QNO có PMO=QNO=90 độ (gt) và POM=QON (2 góc đối đỉnh)

=>tam giác PMO đồng dạng QNO =>PM=QN=x=2,5 =)

Bài 1:

1: ĐKXĐ: x∈R

2: ĐKXĐ: x∈R

3: ĐKXĐ: x-5<>0

=>x<>5

4: ĐKXĐ: 3x+6<>0

=>3x<>-6

=>x<>-2

5: ĐKXĐ: (x+1)(x-3)<>0

=>\(\begin{cases}x+1<>0\\ x-3<>0\end{cases}\Rightarrow\begin{cases}x<>-1\\ x<>3\end{cases}\)

6: ĐKXĐ: \(4-x^2<>0\)

=>\(x^2<>4\)

=>x∉{2;-2}

Câu 4:

a: \(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)

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

a: Xét ΔABC có

AM là đường trung tuyến

G là trọng tâm

Do đó: \(\dfrac{AG}{AM}=\dfrac{2}{3}\)

Xét ΔABM có DG//BM

nên \(\dfrac{AD}{AB}=\dfrac{AG}{AM}\)

=>\(\dfrac{AD}{AB}=\dfrac{2}{3}\)

b: Xét ΔAMC có GE//MC

nên \(\dfrac{AE}{AC}=\dfrac{AG}{AM}\)

=>\(\dfrac{AE}{AC}=\dfrac{2}{3}\)

=>\(AE=\dfrac{2}{3}AC\)

AE+EC=AC

=>\(EC+\dfrac{2}{3}AC=AC\)

=>\(EC=\dfrac{1}{3}AC\)

\(AE=\dfrac{2}{3}AC=2\cdot\dfrac{1}{3}\cdot AC=2\cdot EC\)

Câu 5:

a. $|x+\frac{4}{5}|-\frac{1}{7}=0$

$|x+\frac{4}{5}|=\frac{1}{7}$

$\Rightarrow x+\frac{4}{5}=\pm \frac{1}{7}$

$\Rightarrow x=\frac{-23}{35}$ hoặc $x=\frac{-33}{35}$

v.

$2x+5-(x-7)=18$

$2x+5-x+7=18$

$x+12=18$

$x=6$

c.

$2(x+1)+4^2=2^4$
$2(x+1)+16=16$

$2(x+1)=0$

$x+1=0$

$x=-1$

d.

$\frac{x-3}{x+5}=\frac{5}{7}$

$\Rightarrow 7(x-3)=5(x+5)$

$\Rightarrow 7x-21=5x+25$

$\Rightarrow 2x=46$

$\Rightarrow x=23$

Câu 5:

\(a,\left|x+\dfrac{4}{5}\right|-\dfrac{1}{7}=0\\ \Leftrightarrow\left|x+\dfrac{4}{5}\right|=\dfrac{1}{7}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{4}{5}=\dfrac{1}{7}\\x+\dfrac{4}{5}=-\dfrac{1}{7}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{7}-\dfrac{4}{5}\\x=-\dfrac{1}{7}-\dfrac{4}{5}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{23}{35}\\x=-\dfrac{33}{35}\end{matrix}\right.\\ b,2x+5-\left(x-7\right)=18\\ \Leftrightarrow2x-x=18-5-7\\ \Leftrightarrow x=6\\ c,2\left(x+1\right)+4^2=2^4\\ \Leftrightarrow2\left(x+1\right)=2^4-4^2=16-16\\ \Leftrightarrow2\left(x+1\right)=0\\ \Rightarrow x+1=0\\ \Leftrightarrow x=0-1=-1\\ d,\dfrac{x-3}{x+5}=\dfrac{5}{7}\left(x\ne-5\right)\\ \Leftrightarrow7\left(x-3\right)=5\left(x+5\right)\\ \Leftrightarrow7x-21=5x+25\\ \Leftrightarrow7x-5x=25+21\\ \Leftrightarrow2x=46\\ \Leftrightarrow x=23\)