5 x + 1 - 25 x = 375
Tìm x
\(\dfrac{29-x}{21}\)+\(\dfrac{27-x}{23}\)+\(\dfrac{25-x}{25}\)+\(\dfrac{23-x}{27}\)+\(\dfrac{21-x}{29}\)=\(\dfrac{(29-x+1}{21}\)+\(\dfrac{(27-x+1)}{23}\)+\(\dfrac{(25-x+1)}{25}\)+\(\dfrac{(23-x+1)}{21}\)=-5 +5
GIẢI nốt hộ mình với ạ
1) 444 x 5 = 222 x 2 x 5 = 222 x 10 = 2220
a.Đúng b.Sai
2) 282 x 5 = 280 + 2 x 5 = 280 x 10 = 2800
a.Đúng b. Sai
3) 4 x 8 x 7 x 25 = (8 x 7) x (25 x4) = 56 x 100 = 5600
a.Đúng b.Sai
4) 25 x 8 x 9 = (25 x 4) x (4 x 9) = 100 x 36 = 3600
a.Đúng b.Sai
1. 444 x 5 = 222 x 2 x 5 = 222 x 10 = 2220
2. 282 x 5 = 280 + 2 x 5 = 280 x 10 = 2800
3. 4 x 8 x 7 x 25 = (8 x 7) x (25 x4) = 56 x 100 = 5600
4. 25 x 8 x 9 = (25 x 4) x (4 x 9) = 100 x 36 = 3600
a.Đúng b.Sai
Cho biểu thức: A = x+3√x/x-25 + 1/√x+5; B = √x-5/√x+2 (điều kiện: x ≥ 0, x ≠ 25). P = √x-1/√x+2
Tìm x để P > 1/3
P>1/3
=>P-1/3>0
=>\(\dfrac{\sqrt{x}-1}{\sqrt{x}+2}-\dfrac{1}{3}>0\)
=>\(\dfrac{3\sqrt{x}-3-\sqrt{x}-2}{3\left(\sqrt{x}+2\right)}>0\)
=>2 căn x-5>0
=>x>25/4
các bạn ơi trả lời nhanh câu này . Mình mới làm 1 nửa
25 x 5 x 4 x 125x 20 x 8
25 x 5 x 4 x 25 x 5 x 20 x 8
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thừa các bạn bỏ nhé!
\(25.5.4.125.20.8\)
\(=\left(25.4\right).\left(5.20\right).\left(125.8\right)\)
\(=100.100.1000\)
\(=10000000\)
25*(5+5)*4*20*8
=250*20+4+8
= 5000*12
=60000
|x+25|+|−y+5|=0
⇒|x+25|=0 và |−y+5|=0
+) |x+25|=0
⇒x+25=0
⇒x=−25
+) |−y+5|=0
⇒−y+5=0
⇒−y=−5
⇒y=5
Vậy cặp số (x;y) là (−25;5)
Những câu b-f thì chia ra làm 2 vế rồi tính
g thì tìm ước rồi lập bảng trường hợp trong ước
h. (2x−1).(4y−2)=−42(2x−1).(4y−2)=−42
⇒{2x−1∈Ư(−42)4y−2∈Ư(−42)⇒{2x−1∈Ư(−42)4y−2∈Ư(−42)
Mà: Ư(−42)∈{±1;±2;±3;±6;±7;±21;±42}Ư(−42)∈{±1;±2;±3;±6;±7;±21;±42}
Ta có một số trường hợp sau :
2x−12x−1 | 1 | -1 | 2 | -2 | 3 | -3 | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
(4y−2)=2(2y−1)(4y−2)=2(2y−1) | -1 | 1 | -2 | 2 | -|x+25|+|−y+5|=0 ⇒|x+25|=0 và |−y+5|=0 +) |x+25|=0 ⇒x+25=0 ⇒x=−25 +) |−y+5|=0 ⇒−y+5=0 ⇒−y=−5 ⇒y=5 Vậy cặp số (x;y) là (−25;5)
Những câu b-f thì chia ra làm 2 vế rồi tính g thì tìm ước rồi lập bảng trường hợp trong ước
h. (2x−1).(4y−2)=−42(2x−1).(4y−2)=−42 ⇒{2x−1∈Ư(−42)4y−2∈Ư(−42)⇒{2x−1∈Ư(−42)4y−2∈Ư(−42) Mà: Ư(−42)∈{±1;±2;±3;±6;±7;±21;±42}Ư(−42)∈{±1;±2;±3;±6;±7;±21;±42} Ta có một số trường hợp sau :
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Rút gọn biểu thức chứa chữ A = (1/√x -1 + 1/√x +1 ) : 1/√x -1 với x lớn hơn hoặc bằng 0 , x khác 1 B = 2√x /√x -5 - x -25√x / 25 -x với lớn hơn hoặc bằng 0 , x khác 25
\(A=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right):\dfrac{1}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}+1+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{1}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}+1}\)
(\(\dfrac{x^2-5x}{x^2-25}\)-1):(\(\dfrac{25-x^2}{x^2+2x-15}\)-\(\dfrac{x+3}{x+5}\)-\(\dfrac{x-3}{x-5}\))
1/ (\(\left(-\dfrac{2}{3}\right)\)\(^2\) x \(\dfrac{-9}{8}\) - 25% x \(\dfrac{-16}{5}\)
2/ -1\(\dfrac{2}{5}\) x 75% + \(\dfrac{-7}{5}\) x 25%
3/ -2\(\dfrac{3}{7}\) x (-125%) + \(\dfrac{-17}{7}\) x 25%
4/ (-2)\(^3\) x (\(\dfrac{3}{4}\) x 0.25) : (2\(\dfrac{1}{4}\) - 1\(\dfrac{1}{6}\))
1) Ta có: \(\left(-\dfrac{2}{3}\right)^2\cdot\dfrac{-9}{8}-25\%\cdot\dfrac{-16}{5}\)
\(=\dfrac{4}{9}\cdot\dfrac{-9}{8}-\dfrac{1}{4}\cdot\dfrac{-16}{5}\)
\(=\dfrac{-1}{2}+\dfrac{4}{5}\)
\(=\dfrac{-5}{10}+\dfrac{8}{10}=\dfrac{3}{10}\)
2) Ta có: \(-1\dfrac{2}{5}\cdot75\%+\dfrac{-7}{5}\cdot25\%\)
\(=\dfrac{-7}{5}\cdot\dfrac{3}{4}+\dfrac{-7}{5}\cdot\dfrac{1}{4}\)
\(=\dfrac{-7}{5}\left(\dfrac{3}{4}+\dfrac{1}{4}\right)=-\dfrac{7}{5}\)
3) Ta có: \(-2\dfrac{3}{7}\cdot\left(-125\%\right)+\dfrac{-17}{7}\cdot25\%\)
\(=\dfrac{-17}{7}\cdot\dfrac{-5}{4}+\dfrac{-17}{7}\cdot\dfrac{1}{4}\)
\(=\dfrac{-17}{7}\cdot\left(\dfrac{-5}{4}+\dfrac{1}{4}\right)\)
\(=\dfrac{17}{7}\)
4) Ta có: \(\left(-2\right)^3\cdot\left(\dfrac{3}{4}\cdot0.25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)\)
\(=\left(-8\right)\cdot\left(\dfrac{3}{4}\cdot\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=\left(-8\right)\cdot\dfrac{3}{16}:\dfrac{54-28}{24}\)
\(=\dfrac{-3}{2}\cdot\dfrac{24}{26}\)
\(=\dfrac{-72}{52}=\dfrac{-18}{13}\)
\(\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}+\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
đK: \(x\ge0;x\ne25;x\ne9\)
\(=\left[\dfrac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-1\right]:\left[\dfrac{25-x}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}\right]\)
\(=\left[\dfrac{\sqrt{x}}{\sqrt{x}+5}-1\right]:\dfrac{25-x-\left(x-9\right)+\left(x-25\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-5}{\sqrt{x}+5}:\dfrac{9-x}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-5}{\sqrt{x}+5}:\dfrac{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)\(=\dfrac{-5}{\sqrt{x}+5}:\dfrac{-\sqrt{x}-3}{\sqrt{x}+5}\)
\(=\dfrac{-5}{\sqrt{x}+5}:\dfrac{\sqrt{x}+5}{-\left(\sqrt{x}+3\right)}=\dfrac{5}{\sqrt{x}+3}\)
Rút gọn: \(\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}+\dfrac{\sqrt{x}-5}{\sqrt{ }-3}\)
Ta có: \(\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}+\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
\(=\dfrac{x-5\sqrt{x}-x+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}:\dfrac{25-x-x+9+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-5\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\cdot\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{5}{\sqrt{x}+3}\)