1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 = 362880

= 362880

362880

a) \(\sqrt{x^4}=2\)( ĐK x ∈ R )

⇔ \(\sqrt{\left(x^2\right)^2}=2\)

⇔ \(\left|x^2\right|=2\)

⇔ \(\orbr{\begin{cases}x^2=2\\x^2=-2\left(loai\right)\end{cases}}\)

⇔ x² - 2 = 0

⇔ ( x - √2 )( x + √2 ) = 0

⇔ x - √2 = 0 hoặc x + √2 = 0

⇔ x = ±√2 

b) \(3\sqrt{x+1}-8=0\)( ĐK x ≥ -1 )

⇔ \(3\sqrt{x+1}=8\)

⇔ \(\sqrt{x+1}=\frac{8}{3}\)

⇔ \(x+1=\frac{64}{9}\)

⇔ \(x=\frac{55}{9}\)( tm )

c) \(2\sqrt{x-3}+\sqrt{25x-75}=14\)( ĐK x ≥ 3 ) ( Vầy hợp lí hơn á )

⇔ \(2\sqrt{x-3}+\sqrt{5^2\left(x-3\right)}=14\)

⇔ \(2\sqrt{x-3}+5\sqrt{x-3}=14\)

⇔ \(7\sqrt{x-3}=14\)

⇔ \(\sqrt{x-3}=2\)

⇔ \(x-3=4\)

⇔ \(x=7\)( tm )

d) \(\sqrt{\left(3x-1\right)^2}=5\)( ĐK x ∈ R )

⇔ \(\left|3x-1\right|=5\)

⇔ \(\orbr{\begin{cases}3x-1=5\\3x-1=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-\frac{4}{3}\end{cases}}\)

e) \(\sqrt{x^2+4x+4}-6=0\)( ĐK x ∈ R )

⇔ \(\sqrt{\left(x+2\right)^2}=6\)

⇔ \(\left|x+2\right|=6\)

⇔ \(\orbr{\begin{cases}x+2=6\\x+2=-6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=-8\end{cases}}\)

\(a)\)\(\sqrt{x^4}=2\)\(\Leftrightarrow\)\(x^2=2\)\(\Rightarrow\)\(\orbr{\begin{cases}x=\sqrt{2}\\x=-\sqrt{2}\end{cases}}\)

Vậy \(x=\sqrt{2}\)\(hoặc\)\(x=-\sqrt{2}\)

​\(b)\)\(ĐK:x\ge0\)

\(3\sqrt{x+1}-8=0\)\(\Leftrightarrow\)\(3\sqrt{x}=8\)\(\Leftrightarrow\)\(\sqrt{x}=\frac{8}{3}\)\(\Leftrightarrow\)\(x=(\frac{8}{3})^2\)\(\Leftrightarrow\)\(x=\frac{64}{9}\)\((TM)\)

Vậy \(x=\frac{64}{9}\)

\(d)\)\(\sqrt{(3x-1)^2}=5\)\(\Leftrightarrow\)\(|3x-1|=5\)\((1)\)

Vậy \(x\in\hept{2;\frac{-4}{3}}\)

                -Nếu \(x\ge-2\)thì \(\left(2\right)\Leftrightarrow x+2=6\Leftrightarrow x=4(TM)\)

                -Nếu \(x< -2\)thì \(\left(2\right)\Leftrightarrow-\left(x+2\right)=6\Leftrightarrow x+2=-6\Leftrightarrow x=-8\left(TM\right)\)

Vậy \(x=4;x=-8\)

=1

=4

=9

=16

=25

=36

=42

=64

=81

=100

1

4

9

16

25

36

49

64

81

100

= 1

= 4

= 9

= 16

= 25

= 36

= 49

= 64

= 81

= 100

1/2 * 2/3 * 3/4 * 4*5=1*2*3*4/2*3*4=1/4 ngu , thế mà cũng gọi là toán lớp 5 

1/2*2/3*3/4*4/5=1*2*3*4/2*3*4*5=1/5 ngu, thế mà gọi là toán lớp 5

\(\left[1-\frac{1}{2}\right]x\left[1-\frac{1}{3}\right]x\left[1-\frac{1}{4}\right]x\left[1-\frac{1}{5}\right].\)

\(=\left[\frac{2}{2}-\frac{1}{2}\right]x\left[\frac{3}{3}-\frac{1}{3}\right]x\left[\frac{4}{4}-\frac{1}{4}\right]x\left[\frac{5}{5}-\frac{1}{5}\right].\)

\(=\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x\frac{4}{5}.\)  (Gạch đi những số: 2, 3, 4) (Còn lại 1 và 5)

\(=\frac{1}{5}\)

b: \(\Leftrightarrow2\left(x^2-2x+1\right)-3x^2+5x-1=0\)

\(\Leftrightarrow2x^2-4x+2-3x^2+5x-1=0\)

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

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

\(\text{Δ}=\left(-1\right)^2-4\cdot1\cdot\left(-1\right)=5\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{1-\sqrt{5}}{2}\\x_2=\dfrac{1+\sqrt{5}}{2}\end{matrix}\right.\)

c: \(\Leftrightarrow x^2+6x+9-1-\left(x^2+8x-4x-32\right)=0\)

\(\Leftrightarrow x^2+6x+8-x^2-4x+32=0\)

=>2x+40=0

hay x=-20

d: \(\Leftrightarrow3x^2+12x+12+4x^2-4x+1-7\left(x^2-9\right)=36\)

\(\Leftrightarrow7x^2+8x+13-7x^2+63=36\)

=>8x+76=36

hay x=-5

\(x-\frac{3}{5}=\frac{4}{7}\)                             \(x+\frac{3}{5}=\frac{4}{3}\)                          \(-x-\frac{2}{7}=-\frac{8}{9}\)

\(x=\frac{4}{7}+\frac{3}{5}\)                           \(x=\frac{4}{3}-\frac{3}{5}\)                              \(-x=-\frac{8}{9}+\frac{2}{7}\)

\(x=\frac{41}{35}\)                                        \(x=\frac{11}{15}\)                                      \(-x=-\frac{38}{63}\)

                                                                                                                      \(x=\frac{38}{63}\)

\(\frac{7}{9}-x=\frac{1}{5}\)

\(x=\frac{7}{9}-\frac{1}{5}\)

\(x=\frac{26}{45}\)

cộng ko có đáp án à 

(x + 1 ) + ( x + 2 ) + ( x + 3 ) + ........+ (x + 100) = 5550

x = 100 + ( 1 + 2 + 3  + .........+100)=5550

x = (5550 - 5050 ) : 100

x = 500 : 100

x = 5

lethanhthao em lấy đâu ra 5550 đầu bài ko có bị thiếu