√25x-275 - √9x-99 - √x-11 = 1
Bài 1: Tìm x, biết
a)\(2\sqrt{9x-27}-\dfrac{1}{5}\sqrt{25x-75}-\dfrac{1}{7}\sqrt{49x-147}=20\)
b) \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)
c)\(\sqrt{16x-16}-\sqrt{9x-9}+\sqrt{4x-4}+\sqrt{x-1}=8\)
d) \(\sqrt{x+2\sqrt{x-1}}-\sqrt{x-2\sqrt{x-1}}=2\)
a) Ta có: \(2\sqrt{9x-27}-\dfrac{1}{5}\sqrt{25x-75}-\dfrac{1}{7}\sqrt{49x-147}=20\)
\(\Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\)
\(\Leftrightarrow4\sqrt{x-3}=20\)
\(\Leftrightarrow x-3=25\)
hay x=28
b) Ta có: \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)
\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)
\(\Leftrightarrow2\sqrt{x+2}=6\)
\(\Leftrightarrow x+2=9\)
hay x=7
1/ 2,4 x 1994 x 2+1,6 x 3996 x 3+1,2 x 4010 x 4
2/ 3+7+11+15+...+95+99-275
1. 2,4 x 1994 x 2 + 1,6 x 3996 x 3 + 1,2 x 4010 x 4
= ( 2,4 x 1994 x 2 ) + ( 1,6 x 3996 x 3 ) + ( 1,2 x 4010 x 4 )
= ( 2,4 x 2 x 1994 ) + ( 1,6 x 3 x 3996 ) + ( 1,2 x 4 x 4010 )
= 4,8 x 1994 + 4,8 x 3996 + 4,8 x 4010
= 4,8 x ( 1994 + 3996 + 4010 )
= 4,8 x 10 000
= 48000
2. 3 + 7 + 11 + 15 +........+ 95 + 99 - 275
= [ ( 99 - 3 ) : 4 + 1 ] - 275
= 25 - 275
= [ ( 99 + 3 ) x 25 : 2 ] - 275
= 1275 - 275
= 1000
kick mình nhé bạn
1/ 2,4 x 1994 x 2+1,6 x 3996 x 3+1,2 x 4010 x 4
2/ 3+7+11+15+...+95+99-275
mik cũng có câu hỏi như thế này ,các bn giải thik giúp mik đc ko
Tính nhanh
1/ 2,4 x 1994 x 2+1,6 x 3996 x 3+1,2 x 4010 x 4
2/ 3+7+11+15+...+95+99-275
tinh nhanh 2 bai nay giup mk di ma cac ban oi
tìm x 275% : x - 3/4 : x + 1,2 : x = 1,6 giúp mik với, ai làm đg mik tick ng đó!!!
Tìm x,biết:
a)(1-3x)2-9x(1+x)=-29
b)(2x-1)3-(x-2)2=x(4-25x)-6
\(a,\Rightarrow1-6x+9x^2-9x-9x^2=-29\\ \Rightarrow-15x=-30\Rightarrow x=2\\ b,\Rightarrow8x^3-12x^2+6x-1-x^2+4x-4=4x-25x^2-6\\ \Rightarrow8x^3+12x^2+6x+1=0\\ \Rightarrow\left(2x+1\right)^3=0\\ \Rightarrow2x+1=0\Rightarrow x=-\dfrac{1}{2}\)
12.129+6.437.2+32.364.4/3+7+11+15+......+99-275
\(\sqrt{x-5}+\dfrac{1}{3}\sqrt{9x-45}=\dfrac{1}{5}\sqrt{25x-125+6}\)
\(ĐK:x\ge5\\ \Leftrightarrow\sqrt{x-5}+\dfrac{1}{3}\cdot3\sqrt{x-5}=\dfrac{1}{5}\sqrt{25x-119}\\ \Leftrightarrow2\sqrt{x-5}=\dfrac{1}{5}\sqrt{25x-119}\\ \Leftrightarrow4\left(x-5\right)=\dfrac{1}{25}\left(25x-119\right)\\ \Leftrightarrow4x-20=x-\dfrac{119}{25}\\ \Leftrightarrow3x=\dfrac{381}{25}\Leftrightarrow x=\dfrac{127}{25}\)
Ta có: \(\sqrt{x-5}+\dfrac{1}{3}\sqrt{9x-45}=\dfrac{1}{5}\sqrt{25x-125}+6\)
\(\Leftrightarrow x-5=36\)
hay x=41
tìm x thuộc n ,biết
1, 538-x=275
2,45-9x=18
3,(5x-9):3=12
`1,`
`538 - x = 275`
`\Rightarrow x = 538 - 275`
`\Rightarrow x = 263`
Vậy, `x = 263`
`2,`
`45 - 9x = 18`
`\Rightarrow 9x = 45 - 18`
`\Rightarrow 9x = 27`
`\Rightarrow x = 27 \div 9`
`\Rightarrow x = 3`
Vậy, `x = 3`
`3,`
`(5x - 9) \div 3 = 12`
`\Rightarrow 5x - 9 = 12. 3`
`\Rightarrow 5x - 9 = 36`
`\Rightarrow 5x = 36 + 9`
`\Rightarrow 5x = 45`
`\Rightarrow x = 45 \div 5`
`\Rightarrow x = 9`
Vậy, `x = 9.`
1) \(538-x=275\)
\(x=538-275\)
\(x=263\)
2) \(45-9x=18\)
\(9x=27\)
\(x=3\)
3) \(\left(5x-9\right)\div3=12\)
\(\left(5x-9\right)=12\times3\)
\(5x-9=36\)
\(5x=45\)
\(x=9\)
1. \(538-x=275\)
\(x=538-275\)
\(x=263\)
2.\(45-9x=18\)
\(9x=45-18\)
\(9x=27\)
\(x=27:9\)
\(x=3\)
\(\sqrt{25x+25}-\sqrt{16x+16}+\sqrt{9x+9}-\sqrt{4x+4}+\sqrt{x+1}=27\)
ĐKXĐ: \(x\ge-1\)
\(\sqrt{25\left(x+1\right)}-\sqrt{16\left(x+1\right)}+\sqrt{9\left(x+1\right)}-\sqrt{4\left(x+1\right)}+\sqrt{x+1}=27\)
\(\Leftrightarrow5\sqrt{x+1}-4\sqrt{x+1}+3\sqrt{x+1}-2\sqrt{x+1}+\sqrt{x+1}=27\)
\(\Leftrightarrow3\sqrt{x+1}=27\)
\(\Leftrightarrow\sqrt{x+1}=9\)
\(\Rightarrow x+1=81\)
\(\Rightarrow x=80\) (thỏa mãn)