\(a)2^x+2^{x+3}=144\\ 2^x\times\left(1+2^3\right)=144\\ 2^x\times9=144\\ 2^x=144:9\\ 2^x=16\\ 2^x=2^4\\ \Rightarrow x=4\)

\(b)\left(4x-1\right)^2=25\times9\\ \left(4x-1\right)^2=\left(5\times3\right)^2\\ \Rightarrow4x-1=\pm15\\ \Rightarrow\left[{}\begin{matrix}4x=16\\4x=-14\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{7}{2}\end{matrix}\right.\)

mọi người giúo mik vs ạ

`12m6dm = 126dm`

`5m50cm =55dm`

Diện tích xung quanh :

`55 xx 2 xx ( 126 + 80) = 22660(dm^2)`

Diện tích `1` mặt :

`126 xx 80 =10080(dm^2)`

Diện tích toàn phần :

`22660 + 10080 = 32740(dm^2)`

Đ/s...

\(\overline{3a7b}\) ⋮ cả 5 và 9 ⇔ b= 0 hoặc 5 và 3 + a + 7 + b ⋮ 9

nếu b = 0 ⇔ 3 + 0 + 7 + a ⋮ 9 ⇔ 1 + a ⋮ 9 ⇔ a = 8

 số đó là 3870

nếu b  = 5 ⇔ 3 + a + 7 + 5 ⋮ 9 ⇔ a+ 6 ⋮ 9 ⇔ a = 3

số đó là 3375

các số thóa mãn đề bài là 3870 và 3375

b, tuổi cha là 70: (1 + 0,4)  = 50 (tuổi)

tuổi con là 50 x 0,4 = 20 (tuổi)

đs

đs....

Chi mua số hết số tiền :

`100000 xx 3/10 = 30000 (đồng)`

Chi mua sách hết số tiền :

`100000 xx 1/4 = 25000 (đồng)`

Số tiền còn lại :

`100000 - 30000 - 25000 = 45000 (đồng)`

Đ/s...

Chi mua vở hết bn tiền: 

100.000 x 3/10 = 30.000 (đồng)

Chi mua sách hết bn tiền:

100.000 × 1/4 =25.000 (đồng)

Chi còn lại bn tiền:

100.000 -30.000-25.000 =45.000(đồng)

a) We have \(3\le x\le5\)

Suppose \(\sqrt{5-x}=a\left(a\ge0\right);\sqrt{x-3}=b\left(b\ge0\right)\)

We imidiately have \(a+b=\sqrt{2}\)

On the other hand, we have \(a^2-b^2=\left(\sqrt{5-x}\right)^2-\left(\sqrt{x-3}\right)^2=\left(5-x\right)-\left(x-3\right)=8\)

\(\Leftrightarrow\left(a-b\right)\left(a+b\right)=8\) \(\Leftrightarrow\sqrt{2}\left(a-b\right)=8\) \(\Leftrightarrow a-b=4\sqrt{2}\)

And now we simply have \(\left\{{}\begin{matrix}a+b=\sqrt{2}\\a-b=4\sqrt{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{5\sqrt{2}}{2}\left(take\right)\\b=\dfrac{3\sqrt{2}}{2}\left(take\right)\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{5-x}=\dfrac{5\sqrt{2}}{2}\\\sqrt{x-3}=\dfrac{3\sqrt{2}}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5-x=\dfrac{25}{2}\\x-3=\dfrac{9}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{15}{2}\left(eliminate\right)\\x=\dfrac{15}{2}\left(take\right)\end{matrix}\right.\)

Therefore, this equation have the root \(x=\dfrac{15}{2}\)

b) We have \(x\ge2\)

\(\sqrt{x^2-4}=2\sqrt{x-2}\Leftrightarrow\sqrt{\left(x-2\right)\left(x+2\right)}-2\sqrt{x-2}=0\) \(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x+2}-2\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=0\\\sqrt{x+2}-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=4\end{matrix}\right.\Leftrightarrow x=2\left(take\right)\)

So this equation has the root \(x=2\)

c) We have \(x\ge-\dfrac{4}{3}\). Suppose \(2x+3=A\left(A\ge0\right);3x+4=B\left(B\ge0\right)\). Notice that \(A+B=2x+3+3x+4=5x+7\). Thus, we can rewrite the equation as below:

\(\sqrt{A+B}=\sqrt{A}+\sqrt{B}\) (1)

Remember that \(\sqrt{A}+\sqrt{B}\ge\sqrt{A+B}\left(A,B\ge0\right)\). We can prove it easily by squaring each side of this inequality:

\(\sqrt{A}+\sqrt{B}\ge\sqrt{A+B}\Leftrightarrow\left(\sqrt{A}+\sqrt{B}\right)^2\ge\left(\sqrt{A+B}\right)^2\)\(\Leftrightarrow A+B+2\sqrt{AB}\ge A+B\Leftrightarrow2\sqrt{AB}\ge0\Leftrightarrow\sqrt{AB}\ge0\). This is always true when \(A,B\ge0\). "=" happens when one of the expression A, B is equal to 0.

Therefore, (1) happens when \(\left[{}\begin{matrix}A=0\\B=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\left(eliminate\right)\\x=-\dfrac{4}{3}\left(take\right)\end{matrix}\right.\)

Thus, this equation has the root \(x=-\dfrac{4}{3}\)

d) We have \(x\ge-\dfrac{1}{3}\)

We can see that \(x=-\dfrac{1}{11}\). So, \(\sqrt{5x+7}=\dfrac{6\sqrt{22}}{11}\); o\(\sqrt{x+3}=\dfrac{4\sqrt{22}}{11}\) and \(\sqrt{3x+1}=\dfrac{2\sqrt{22}}{11}\). Therefore, we can rewrite the equation as below:

\(\left(\sqrt{5x+7}-\dfrac{6\sqrt{22}}{11}\right)-\left(\sqrt{x+3}-\dfrac{4\sqrt{22}}{11}\right)-\left(\sqrt{3x+1}-\dfrac{2\sqrt{22}}{11}\right)=0\)Now you multiply and devide each of the terms by the conjugate expression. The first term will has \(5x+\dfrac{5}{11}=5\left(x+\dfrac{1}{11}\right)\) as the numerator, the second term's numerator will be \(x+\dfrac{1}{11}\), and the final term has \(3x+\dfrac{3}{11}=3\left(x+\dfrac{1}{11}\right)\) as the numerator. And now you can see the commom factor \(x+\dfrac{1}{11}\)

\(\left(\sqrt{A}-B=\dfrac{\left(\sqrt{A}-B\right)\left(\sqrt{A}+B\right)}{\sqrt{A}+B}=\dfrac{A-B^2}{\sqrt{A}-B}\right)\)

789 = 7.10² + 8.10¹ +9.10⁰

25436 = 2.10⁴  + 5.10³ +4.10² +3.10¹ +6.10⁰

Giúp mình với, mình cần gấp

=) S = -2 + ( 4 - 6 ) + ... + ( 24-26)+28

=) S = -2 + (-2) + ( -2) +...+(-2)+28

=)S=-2x8+28

=)S=-16+28=12

S = -2 + 4 - 6 + 8 -......-26+28

S = 28 -26 + 24 - 22 + .......+ 4 -2 

S = (28 - 26) + ( 24 - 22) +.....+ (4-2)

tổng trên có số nhóm là (28 - 4) : 4 + 1 = 7 nhóm,

mỗi nhóm có giá trị là      28 - 26 = 2

S =  2 x 7 = 14

ok babe

x + \(\dfrac{x}{3}\) = 24

\(\dfrac{x\times3}{3}\) + \(\dfrac{x}{3}\) = \(\dfrac{72}{3}\)

3x \(x\) + \(x\) = 72

\(x\) x (3+1) = 72

\(x\) x 4 = 72

\(x\) = 72: 4

\(x\) = 18