Bài 4 : Tính nhanh :
a, 15. 64 + 25. 100 + 36. 15 + 60. 100

= (15 . 64 + 36. 15) + (25. 100 + 60. 100)

= 15.(64 + 36) + 100.(25 + 60)

= 15. 100 + 100. 85

= 100.(15 + 85)

= 100. 100

= 10000
b, 47² + 48² - 25 + 94. 48

= 47² + 2.47. 48 + 48² - 25

= (47 + 48)² - 5²

= (47 + 48 - 5)(47 + 48 + 5)

= (48 + 22)(48 + 52)

= 90. 100

= 9000
c, 9³ - 9². ( -1) - 9. 11 + ( -1). 11

= 9³ + 9² + 11(- 9 - 1)

= 9².(9 + 1) + 11. (-10)

= 81. 10 - 110

= 810 - 110

= 700
d,2016. 2018 - 2017²

= (2017 - 1)(2017 + 1) - 2017²

= 2017² - 1 - 2017²

= -1

#Học tốt!

các bn lm đến đâu cx dc miễn là lm hộ mk cái ạ, ai đang lm vào nhắn tin vs mk để mk bít nha

a; \(-\dfrac{8}{3}+\dfrac{7}{5}-\dfrac{71}{15}< x< -\dfrac{13}{7}+\dfrac{19}{14}-\dfrac{7}{2}\)

              -\(\dfrac{19}{15}\) - \(\dfrac{71}{15}\) < \(x\) < -\(\dfrac{1}{2}\) - \(\dfrac{7}{2}\)

              -6 < \(x\) < -4

             vì \(x\) \(\in\) Z nên \(x\) = -5

- \(\dfrac{19}{6}+\dfrac{-15}{2}+\dfrac{11}{3}\) < \(x\) < - \(\dfrac{5}{4}\) + \(\dfrac{19}{12}\) - \(\dfrac{10}{3}\)

- \(\dfrac{32}{3}+\dfrac{11}{3}< x< \) \(\dfrac{1}{3}\) - \(\dfrac{10}{3}\)

-7 < \(x\) < - 3

Vì \(x\in\) Z nên \(x\in\) {-6;-5;-4}

6/20:3=6/20x1/3=6/60=1/10

11h-8h30p=2h30p=2.5h

10 7/10-4 3/10

=107/10-43/10

=64/10=32/5

chúc bn học tốt!

\(x+x\cdot3:\dfrac{2}{9}+x:\dfrac{2}{7}=252\)

\(\Leftrightarrow x+x\cdot3\cdot\dfrac{9}{2}+x\cdot\dfrac{7}{2}=252\)

\(\Leftrightarrow x\cdot18=252\)

hay x=14

Lời giải:

a. ĐKXĐ: $x\geq -9$

PT $\Leftrightarrow x+9=7^2=49$

$\Leftrightarrow x=40$ (tm)

b. ĐKXĐ: $x\geq \frac{-3}{2}$

PT $\Leftrightarrow 4\sqrt{2x+3}-\sqrt{4(2x+3)}+\frac{1}{3}\sqrt{9(2x+3)}=15$

$\Leftrightarrow 4\sqrt{2x+3}-2\sqrt{2x+3}+\sqrt{2x+3}=15$

$\Leftrgihtarrow 3\sqrt{2x+3}=15$

$\Leftrightarrow \sqrt{2x+3}=5$

$\Leftrightarrow 2x+3=25$

$\Leftrightarrow x=11$ (tm)

c.

PT \(\Leftrightarrow \left\{\begin{matrix} 2x+1\geq 0\\ x^2-6x+9=(2x+1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ 3x^2+10x-8=0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ (3x-2)(x+4)=0\end{matrix}\right.\)

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

d. ĐKXĐ: $x\geq 1$

PT \(\Leftrightarrow \sqrt{(x-1)+4\sqrt{x-1}+4}-\sqrt{(x-1)+6\sqrt{x-1}+9}=9\)

\(\Leftrightarrow \sqrt{(\sqrt{x-1}+2)^2}-\sqrt{(\sqrt{x-1}+3)^2}=9\)

\(\Leftrightarrow \sqrt{x-1}+2-(\sqrt{x-1}+3)=9\)

\(\Leftrightarrow -1=9\) (vô lý)

Vậy pt vô nghiệm.

a) \(\sqrt{x+9}=7\left(x\ge-9\right)\Rightarrow x+9=49\Rightarrow x=40\)

b) \(4\sqrt{2x+3}-\sqrt{8x+12}+\dfrac{1}{3}\sqrt{18x+27}=15\left(x\ge-\dfrac{3}{2}\right)\)

\(\Rightarrow4\sqrt{2x+3}-\sqrt{4\left(2x+3\right)}+\dfrac{1}{3}\sqrt{9\left(2x+3\right)}=15\)

\(\Rightarrow4\sqrt{2x+3}-2\sqrt{2x+3}+\sqrt{2x+3}=15\)

\(\Rightarrow3\sqrt{2x+3}=15\Rightarrow\sqrt{2x+3}=5\Rightarrow2x+3=25\Rightarrow x=11\)

c) \(\sqrt{x^2-6x+9}=2x+1\)

Vì \(VT\ge0\Rightarrow VP\ge0\Rightarrow x\ge-\dfrac{1}{2}\)

\(\Rightarrow\sqrt{\left(x-3\right)^2}=2x+1\Rightarrow\left|x-3\right|=2x+1\Rightarrow\left[{}\begin{matrix}x-3=2x+1\\x-3=-2x-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-4\left(l\right)\\x=\dfrac{2}{3}\end{matrix}\right.\)

d) \(\sqrt{x+3+4\sqrt{x-1}}-\sqrt{x+8+6\sqrt{x-1}}=9\left(x\ge1\right)\)

\(\Rightarrow\sqrt{x-1+4\sqrt{x-1}+4}-\sqrt{x-1+6\sqrt{x-1}+9}=9\)

\(\Rightarrow\sqrt{\left(\sqrt{x-1}+2\right)^2}-\sqrt{\left(\sqrt{x-1}+3\right)^2}=9\)

\(\Rightarrow\left|\sqrt{x-1}+2\right|-\left|\sqrt{x-1}+3\right|=9\)

\(\Rightarrow\sqrt{x-1}+2-\sqrt{x-1}-3=9\Rightarrow-1=9\) (vô lý)

\(3\cdot2\cdot x-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):\dfrac{11}{3}=\dfrac{7}{20}\)

\(6x-\left(\dfrac{12}{15}+\dfrac{10}{15}\right)\cdot\dfrac{3}{11}=\dfrac{7}{20}\)

\(6x-\dfrac{22}{15}\cdot\dfrac{3}{11}=\dfrac{7}{20}\\ 6x-\dfrac{2}{5}=\dfrac{7}{20}\\ 6x=\dfrac{7}{20}+\dfrac{2}{5}\\ 6x=\dfrac{3}{4}\\ x=\dfrac{1}{8}\)

\(3.2x-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):\dfrac{11}{3}=\dfrac{7}{20}\)

\(\Rightarrow6x-\dfrac{22}{15}:\dfrac{11}{3}=\dfrac{7}{20}\)

\(\Rightarrow6x-\dfrac{2}{5}=\dfrac{7}{20}\)

\(\Rightarrow6x=\dfrac{7}{20}+\dfrac{2}{5}\)

\(\Rightarrow6x=\dfrac{3}{4}\)

\(\Rightarrow x=\dfrac{3}{4}:6\)

\(\Rightarrow x=\dfrac{1}{8}\)

\(a,\left(x-3\right)^2-4=0\)

\(\Leftrightarrow\left(x-3\right)^2=4\)

\(\Rightarrow x-3=\pm2\)

\(\hept{\begin{cases}x-3=2\Rightarrow x=5\\x-3=-2\Rightarrow x=1\end{cases}}\)

Vậy \(x=5\)hoặc \(x=1\)

\(b,x^2-2x=24\)

\(\Leftrightarrow x^2-2x+1-1=24\)

\(\Leftrightarrow\left(x-1\right)^2=24+1=25\)

\(\Leftrightarrow x-1=\pm5\)

\(\hept{\begin{cases}x-1=5\Rightarrow x=6\\x-1=-5\Rightarrow x=-4\end{cases}}\)

Vậy \(x=6\) hoặc \(x=-4\)

\(c,\left(2x+1\right)^2+\left(x+3\right)^2-5\left(x-7\right)\left(x+7\right)=0\)

\(\Leftrightarrow4x^2+4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)

\(\Leftrightarrow4x^2+4x+1+x^2+6x+9-5x^2+245=0\)

\(\Leftrightarrow10x+255=0\)

\(\Leftrightarrow10x=-255\)

\(\Leftrightarrow x=\frac{-51}{2}\)

\(d,\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=1\)

\(\Leftrightarrow x^3-27+x\left(2x-x^2+4-2x\right)=1\)

\(\Leftrightarrow x^3-27-x^3+4x=1\)

\(\Leftrightarrow4x-27=1\)

\(\Leftrightarrow4x=28\)

\(\Leftrightarrow x=7\)

\(e,\left(3x-1\right)^2+2\left(x+3\right)^2+11\left(x+1\right)\left(1-x\right)=6\)

\(\Leftrightarrow9x^2-6x+1+2\left(x^2+6x+9\right)+11\left(x-x^2+x-x\right)=6\)

\(\Leftrightarrow9x^2-6x+1+2x^2+12x+18-11x^2+11x=6\)

\(\Leftrightarrow17x+19=6\)

\(\Leftrightarrow17x=-13\)

\(\Leftrightarrow x=\frac{-13}{17}\)

em ko biết

Có bị lạc đề ko dzậy?!

Phần e, 17-{-x-[-x-(-x)]}=-16 nha mn, chỗ này thiếu:))