Có : abc x 9 = a0bc

( a x 100 + bc ) x 9 = a x 1000 + bc

900 x a + 9 x bc = 1000 x 4 + bc

9 x bc - bc = 1000 x a - 900 x a

8 x bc = 100 x a = > bc = ...... ; a = ....

Ngủ ngon nhé, au bn lém.

are you ok

I don't know

k minh di minh cho cach giai.Bai nay de qua

ns vậy mik cx bt lm nhưng k mik mik cho lời giải,đ là chắc

Ta có sơ đồ :

Số cần tìm : |-----------|

Số đã cho  : |-----------|-----------|-----------|-----------|-----------|-----------|-----------|-----------|-----------|

Theo sơ đồ, hiệu số phần bằng nhau là :

9 - 1 = 8 ( phần )

Số cần tìm là :

1000 : 8 x 1 = 125

Đáp số : 125

\(5x-9=2x+15\)

\(\Leftrightarrow5x-2x=15+9\)

\(\Leftrightarrow3x=24\)

\(\Leftrightarrow x=8\)

\(2\left(x-3\right)-4\left(x+4\right)=3.\left(-7\right)+5\)

\(\Leftrightarrow2x-6-4x-16=-16\)

\(\Leftrightarrow2x-4x-6-16=-16\)

\(\Leftrightarrow-2x=6\)

\(\Leftrightarrow-x=3\)

\(\Leftrightarrow x=-3\)

\(\left|2x+1\right|-3=9\)

\(\left|2x+1\right|=12\)

\(\Rightarrow\orbr{\begin{cases}2x+1=12\\2x+1=-12\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{11}{2}\\x=\frac{-13}{2}\end{cases}}\)

P/s: Hoq chắc

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}

   136 x 68 + 16 x 272

= 136 x 68 + 16 x  2 x 136

= 136 x 68 + 32 x 136

= 136 x ( 68 + 32 )

= 136 x 100

= 13600    

136x68+16x272

=9248 + 4352

=13600

tk cho mik nha

\(1)\frac{1}{5}+\frac{2}{11}< \frac{x}{55}< \frac{2}{5}+\frac{1}{55}\)

\(\Rightarrow\frac{11}{55}+\frac{10}{55}< \frac{x}{55}< \frac{22}{55}+\frac{1}{55}\)

\(\Rightarrow\frac{21}{55}< \frac{x}{55}< \frac{23}{55}\)

\(\Rightarrow21< x< 23\)

\(\Rightarrow x=22\)

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

\(\Rightarrow\frac{22}{6}+\frac{-19}{6}+\frac{-45}{6}\le x\le\frac{19}{12}+\frac{-15}{12}+\frac{-40}{12}\)

\(\Rightarrow\frac{22+\left[-19\right]+\left[-45\right]}{6}\le x\le\frac{19+\left[-15\right]+\left[-40\right]}{12}\)

\(=\frac{-42}{6}\le x\le\frac{-36}{12}\)

\(\Rightarrow-7\le x\le-3\)

\(\Rightarrow x\in\left\{-7;-6;-5;-4;-3\right\}\)

\(3)\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\le x\le\frac{-1}{2}+2+\frac{5}{2}\)

\(\Rightarrow\frac{3}{6}+\frac{2}{6}+\frac{1}{6}\le x\le\frac{-1}{2}+\frac{5}{2}+2\)

\(\Rightarrow1\le x\le2+2\)

\(\Rightarrow1\le x\le4\)

\(\Rightarrow x\in\left\{1;2;3;4\right\}\)

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)

\(xy+y+x=0\)

\(\Rightarrow y\left(x+1\right)+x+1=1\)

\(\Rightarrow\left(x+1\right)\left(y+1\right)=1\cdot1=\left(-1\right)\left(-1\right)\)

lập bảng

Ta có : x+y+xy=0

   x(y+1) + y    = 0

  x(y+1) + y+ 1 =1

  (y+1)(x+1)      = 1

Vì x, y \(\in Z\)

=> x+1; y+1 là ước của 1

Ta có bảng sau:

Vậy x=y=0 hoặc x=y=-2 

k tui nha