D=/x-3/+/x+7/+/x-9/+/x-11/ mk đang cần gấp
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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
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.
\(3\cdot2\cdot x-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):\dfrac{11}{3}=\dfrac{7}{20}\)
\(6x-\left(\dfrac{12}{15}+\dfrac{10}{15}\right)\cdot\dfrac{3}{11}=\dfrac{7}{20}\)
\(6x-\dfrac{22}{15}\cdot\dfrac{3}{11}=\dfrac{7}{20}\\ 6x-\dfrac{2}{5}=\dfrac{7}{20}\\ 6x=\dfrac{7}{20}+\dfrac{2}{5}\\ 6x=\dfrac{3}{4}\\ x=\dfrac{1}{8}\)
\(a,\left(x-3\right)^2-4=0\)
\(\Leftrightarrow\left(x-3\right)^2=4\)
\(\Rightarrow x-3=\pm2\)
\(\hept{\begin{cases}x-3=2\Rightarrow x=5\\x-3=-2\Rightarrow x=1\end{cases}}\)
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 \(x=-4\)
\(c,\left(2x+1\right)^2+\left(x+3\right)^2-5\left(x-7\right)\left(x+7\right)=0\)
\(\Leftrightarrow4x^2+4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)
\(\Leftrightarrow4x^2+4x+1+x^2+6x+9-5x^2+245=0\)
\(\Leftrightarrow10x+255=0\)
\(\Leftrightarrow10x=-255\)
\(\Leftrightarrow x=\frac{-51}{2}\)
\(d,\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=1\)
\(\Leftrightarrow x^3-27+x\left(2x-x^2+4-2x\right)=1\)
\(\Leftrightarrow x^3-27-x^3+4x=1\)
\(\Leftrightarrow4x-27=1\)
\(\Leftrightarrow4x=28\)
\(\Leftrightarrow x=7\)
Đề bài yêu cầu gì?
GTNN của D