\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)

\(ĐK:x\ge\dfrac{3}{2}\)

\(pt\Leftrightarrow4x^2-9=4\left(x+3\right)\)

\(\Leftrightarrow4x^2-9=4x+12\)

\(\Leftrightarrow4x^2-4x-21=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{22}}{2}\left(l\right)\\x=\dfrac{1+\sqrt{22}}{2}\left(tm\right)\end{matrix}\right.\)

\(b)\sqrt{4x-20}+3.\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)

\(ĐK:x\ge5\)

\(pt\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)

\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\)

\(\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)

\(c)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27.\sqrt{\dfrac{x-1}{81}}=4\)

ĐK:x>=1

\(pt\Leftrightarrow2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4\)

\(\Leftrightarrow4\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=1\)

\(\Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\)

\(d)5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)

\(ĐK:x\ge3\)

\(pt\Leftrightarrow3\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)

\(\Leftrightarrow-\dfrac{5}{3}\sqrt{x-3}-\sqrt{x^2-9}=0\Leftrightarrow\dfrac{5}{3}\sqrt{x-3}+\sqrt{x^2-9}=0\)

\(\Leftrightarrow(\dfrac{5}{3}+\sqrt{x+3})\sqrt{x-3}=0\)

\(\Leftrightarrow\sqrt{x-3}=0\)    (vì \(\dfrac{5}{3}+\sqrt{x+3}>0\))

\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(nhận\right)\)

=>\(5\cdot\dfrac{3\sqrt{x-3}}{5}-7\cdot\dfrac{2\sqrt{x-3}}{3}-7\cdot\sqrt{x^2-9}+18\cdot\sqrt{\dfrac{9}{81}\left(x^2-9\right)}=0\)

=>\(3\cdot\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}=7\cdot\sqrt{x^2-9}-18\cdot\dfrac{3}{9}\cdot\sqrt{x^2-9}\)

=>\(-\dfrac{5}{3}\sqrt{x-3}=\sqrt{x^2-9}\)

=>\(\sqrt{x-3}\left(\sqrt{x+3}+\dfrac{5}{3}\right)=0\)

=>x-3=0

=>x=3

\(\dfrac{4}{25}=\dfrac{4\times4}{25\times4}=\dfrac{16}{100}\\ \dfrac{-7}{4}=\dfrac{-7\times25}{4\times25}=\dfrac{-175}{100}\\ \dfrac{9}{50}=\dfrac{9\times2}{50\times2}=\dfrac{18}{100}\)

\(25\sqrt{\dfrac{x-3}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\left(x\ge3\right)\)

\(=25\sqrt{\dfrac{1}{25}.\left(x-3\right)}-7\sqrt{\dfrac{4}{9}.\left(x-3\right)}-7\sqrt{x^2-9}+18\sqrt{\dfrac{1}{9}.\left(x^2-9\right)}=0\)

\(=5\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)

\(\Rightarrow\dfrac{1}{3}\sqrt{x-3}-\sqrt{\left(x-3\right)\left(x+3\right)}=0\Rightarrow\sqrt{x-3}-3\sqrt{\left(x-3\right)\left(x+3\right)}=0\)

\(\Rightarrow\sqrt{x-3}\left(1-3\sqrt{x+3}\right)=0\Rightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\1=3\sqrt{x+3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{26}{9}\left(l\right)\end{matrix}\right.\)

cảm ơn nhaa<33

Số trung bình cộng về thời gian làm một bài toán của 50 học sinh.

Tần số lớn nhất là 9, giá trị ứng với tần số 9 là 8. Vậy Mốt của dấu hiệu: Mo = 8 (phút).

1/2×(2×x-3)+105/2=-137/2

1/2×(2×x-3)=-137/2-105/2

1/2×(2×x-3)=-242/2

2×x-3=-242/2:1/2

2×x-3=-242

2.x=(-242)+3

2.x=239

x=239:2

x=239/2

A,-12×25-25×75-25×13

=[(-12)-75-13].25

=(-100).25

-2500

B,-50/7×49/10-35/2×-10/7+ -25/3× -9/5

=(-35)-(-25)+(-13)

=-23

C,-354+265-156+125

=[(-354)-156]+(265+125)

=(-510)+277

=-233

D,-74+40-50+16-35

=(-74)+40+(-50)+16+(-35)

=[(-74)+(-35)]+[40+(-50)+16]

=(-109)+26

=-83

E,-2/3×4/5+-4/5×4/3-0,125

=-2/3.4/5+4/5.(-4/3)-0,125

=4/5.[-2/3+(-4/3)]-0,125

=4/5.(-2)-0,125

=-8/5-0,125

=(-1,6)+(-0,125)

=-1,725

-2/3×4/5+ -4/5×4/3-0,125

Giúp e nốt ik ạ

a) Vì tổng A có 25 số hạng nên A = \(\dfrac{\left(1+25\right).25}{2}=325\)

b) Số số hạng là:

\(\left(50-2\right):2+1=25\) \(\left(số\right)\)

Tổng là:

\(\left(2+50\right).25:2=650\) 

c) Số số hạng là:

\(\left(51-3\right):2+1=25\) \(\left(số\right)\)

Tổng là:

\(\left(3+51\right).25:2=675\)

d) Số số hạng là:

\(\left(81-1\right):4+1=21\) \(\left(số\right)\)

Tổng là:

\(\left(1+81\right).21:2=861\)

\(#Wendy.Dang\)

Đề bài là gì vậy bnn??