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1) a) Ta có: \(\frac{x}{-15}=\frac{-60}{x}\) \(\Rightarrow x^2=\left(-15\right).\left(-60\right)=900\)
\(\Rightarrow x=30\)
b) \(\frac{-2}{x}=\frac{-x}{\frac{8}{25}}\) \(\Rightarrow x.\left(-x\right)=\left(-2\right).\frac{8}{25}\)
\(\Rightarrow x.\left(-x\right)=\frac{-16}{25}\)
\(\Rightarrow x.\left(-x\right)=\left(\frac{-4}{5}\right).\frac{4}{5}\)
Vậy \(x=\frac{4}{5}\)
2) a) \(3,8: \left(2x\right)=\frac{1}{4}:2\frac{2}{3}\)
\(\Rightarrow3,8: \left(2x\right)=\frac{3}{32}\)
\(\Rightarrow2x=\frac{3}{32}:3,8=\frac{15}{608}\)
\(x=\frac{15}{608}:2=\frac{15}{1216}\)
Vậy \(x=\frac{15}{1216}\)
b) \(\left(0,25x\right):3=\frac{5}{6}:0,125\)
\(\Rightarrow\left(0,25x\right):3=\frac{20}{3}\)
\(\Rightarrow0,25x=\frac{20}{3}.3=20\)
\(\Rightarrow x=20:0,25=80\)
Vậy x = 80
c) \(0,01:2,5=\left(0,75x\right):0,75\)
\(\Rightarrow\frac{1}{250}=\left(0,75x\right):0,75\)
\(\Leftrightarrow0,75x=\frac{1}{250}.0,75=\frac{3}{1000}\)
\(\Rightarrow x=\frac{3}{1000}:0,75=\frac{1}{250}\)
Vậy \(x=\frac{1}{250}\)
d) \(1\frac{1}{3}:0,8=\frac{2}{3}:\left(0,1x\right)\)
\(\Rightarrow\frac{5}{3}=\frac{2}{3}:\left(0,1x\right)\)
\(\Rightarrow0,1x=\frac{5}{3}.\frac{2}{3}=\frac{10}{9}\)
\(\Rightarrow x=\frac{10}{9}:0,1=\frac{100}{9}\)
Vậy \(x=\frac{100}{9}\)
a) \(\frac{x}{-15}=\frac{-60}{x}\Leftrightarrow x.x=-15.\left(-60\right)\Leftrightarrow x^2=900\Leftrightarrow x^2=\orbr{\begin{cases}30^2\\\left(-30\right)^2\end{cases}}\Leftrightarrow x=\orbr{\begin{cases}30\\-30\end{cases}}\)
Bài 1:
a) \(\frac{x}{-15}=\frac{-60}{x}\Rightarrow x^2=\left(-60\right).\left(-15\right)=900\Rightarrow x=\orbr{\begin{cases}30\\-30\end{cases}}\)
Bài 2: Đặt \(\frac{x}{4}=\frac{y}{7}=k\Rightarrow x=4k;y=7k\)
\(\Rightarrow xy=4k.7k=28k^2=112\)
\(\Rightarrow k^2=4\Rightarrow k=\pm2\)
\(\Rightarrow\orbr{\begin{cases}x=4.2=8\\x=-4.2=-8\end{cases}}\)
Và \(\orbr{\begin{cases}y=7.2=14\\y=-7.2=-14\end{cases}}\)
Bài 3: \(1\frac{1}{3}:0,8=\frac{2}{3}:\left(0,1x\right)\)
\(\Rightarrow\frac{4}{3}:\frac{4}{5}=\frac{2}{3}:\frac{1}{10}x\Rightarrow\frac{5}{3}=\frac{2}{3}:\frac{1}{10}x\)
\(\Rightarrow\frac{1}{10}x=\frac{2}{5}\Rightarrow x=4\)
Mk trả lời nốt bài 4 hộ bn MMS_Hồ Khánh Châu nha:
Bài 4:
Gọi x là giá trị chung của 2 phân số trên.
Ta có: \(\frac{a}{b}=\frac{c}{d}=x\)
\(\Rightarrow a=x.b
\)
\(c=x.d\)
Ta lại có:
\(\frac{a+c}{b+d}=\frac{x.b+x.d}{b+d}=\frac{x.\left(b+d\right)}{b+d}=x\)
Và \(\frac{a}{b}=x\)
\(\Rightarrow\frac{a}{b}=\frac{a+c}{b+d}\)
Vậy \(\frac{a}{b}=\frac{a+c}{b+d}\)
Hk tốt nha
a/\(\left(2-x\right)\times-3=\left(3x-1\right)\times4\)4
\(\Rightarrow-6+3x=12x-4\)
\(\Rightarrow-2=9x\)
\(\Rightarrow x=\frac{-2}{9}\)
bài b cx tương tự nha
ta có;\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}=\frac{a+b}{c+d}\)(THEO TÍNH CHẤT DÃY TỈ SỐ BẰNG NHAU)
\(\Rightarrowđpcm\)
Có : a/ab+a+1 = a/ab+a+abc = 1/b+1+bc = 1/bc+b+1
c/ca+c+1 = bc/abc+bc+b = b/1+bc+b = b/bc+b+1
=> A = 1+bc+b/bc+b+1 = 1
Tk mk nha
BÀI 1:
\(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{a\left(bc+b+1\right)}+\frac{abc}{ab\left(ca+c+1\right)}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{abc+ab+a} +\frac{abc}{a^2bc+abc+ab}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{ab+a+1}+\frac{1}{ab+a+1}\) (thay abc = 1)
\(=\frac{a+ab+1}{a+ab+1}=1\)
1) Ta có : \(\frac{2016a+b+c+d}{a}=\frac{a+2016b+c+d}{b}=\frac{a+b+2016c+d}{c}=\frac{a+b+c+2016d}{d}\)
Trừ 4 vế với 2015 ta được : \(\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
Nếu a + b + c + d = 0
=> a + b = -(c + d)
=> b + c = (-a + d)
=> c + d = -(a + b)
=> d + a = (-b + c)
Khi đó M = (-1) + (-1) + (-1) + (-1) = - 4
Nếu a + b + c + d\(\ne0\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}=\frac{1}{d}\Rightarrow a=b=c=d\)
Khi đó M = 1 + 1 + 1 + 1 = 4
2) a) Ta có : \(\hept{\begin{cases}\left|x+2013\right|\ge0\forall x\\\left(3x-7\right)^{2004}\ge0\forall y\end{cases}\Rightarrow\left|x+2013\right|+\left(3x-7\right)^{2014}\ge0}\)
Dấu "=" xảy ra \(\hept{\begin{cases}x+2013=0\\3y-7=0\end{cases}\Rightarrow\hept{\begin{cases}x=-2013\\y=\frac{7}{3}\end{cases}}}\)
b) 72x + 72x + 3 = 344
=> 72x + 72x.73 = 344
=> 72x.(1 + 73) = 344
=> 72x = 1
=> 72x = 70
=> 2x = 0 => x = 0
c) Ta có :
\(\frac{7}{2x+2}=\frac{3}{2y-4}=\frac{5}{x+4}\Leftrightarrow\frac{7}{2x+2}=\frac{3}{2y-4}=\frac{10}{2x+8}=\frac{7-10}{2x+2-2x-8}=\frac{1}{2}\)(dãy tỉ số bằng nhau)
=> 2x + 2 = 14 => x = 6 ;
2y - 4 = 6 => y = 5 ;
6 + 5 + z = 17 => z = 6
Vậy x = 6 ; y = 5 ; z = 6
3) a) Ta có : \(\frac{a+b+c}{a+b-c}=\frac{a-b+c}{a-b-c}=\frac{a+b+c-a+b-c}{a+b-c-a+b+c}=\frac{2b}{2b}=1\)(dãy ti số bằng nhau)
=> a + b + c = a + b - c => a + b + c - a - b + c = 0 => 2c = 0 => c = 0;
Lại có : \(\frac{a+b+c}{a+b-c}-1=\frac{a-b+c}{a-b-c}-1\Leftrightarrow\frac{2c}{a+b-c}=\frac{2c}{a-b-c}\Rightarrow a+b-c=a-b-c\) => b = 0
Vậy c = 0 hoặc b = 0
c) Ta có : \(\frac{a+b}{c}=\frac{b+c}{a}=\frac{a+c}{b}=\frac{a+b+b+c+a+c}{c+a+b}=2\)(dãy tỉ số bằng nhau)
=> \(\hept{\begin{cases}a+b=2c\\b+c=2a\\a+c=2b\end{cases}}\)
Khi đó P = \(\left(1+\frac{c}{b}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{b}{a}\right)=\frac{b+c}{b}.\frac{c+a}{c}=\frac{a+b}{a}=\frac{2a.2b.2c}{abc}=8\)
Vậy P = 8
2. b) \(7^{2x}+7^{2x+3}=344\)
\(7^{2x}\cdot\left(1+7^3\right)=344\)
\(7^{2x}\cdot\left(1+343\right)=344\)
\(7^{2x}\cdot344=344\)
\(7^{2x}=1\)
\(7^{2x}=7^0\)
\(2x=0\)
\(x=0\)
bài 1:
a) \(\frac{x-3}{x+5}=\frac{5}{7}\)
\(\Leftrightarrow7\left(x-3\right)=5\left(x+5\right)\)
\(\Leftrightarrow7x-21=5x+25\)
\(\Leftrightarrow2x=46\)
\(\Leftrightarrow x=23\)
b) \(\frac{7}{x-1}=\frac{x+1}{9}\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)=7\cdot9\)
\(\Leftrightarrow x^2-1=63\)
\(\Leftrightarrow x^2=64\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=8\\x=-8\end{array}\right.\)
c) \(\frac{x+4}{20}=\frac{5}{x+4}\)
\(\Leftrightarrow\left(x+4\right)^2=5\cdot20\)
\(\Leftrightarrow\left(x+4\right)^2=100\)
\(\Leftrightarrow x+4=10\)
\(\Leftrightarrow x=6\)
a) \(\frac{x-3}{x+5}=\frac{5}{7}\) điều kiện x khác -5
<=> 7(x-3)=5(x+5)
<=> 7x-5x=25+21
<=> x=23
vậy x=23
b) \(\frac{7}{x-1}=\frac{x+1}{9}\)điều kiện x khác 1
<=> 63=x2-1<=> x=\(\pm\)8
vậy x={-8;8}
c) \(\frac{x+4}{20}=\frac{5}{x+4}\) điều kiện x khác -4
<=> (x+4)2=25
<=> \(\left[\begin{array}{nghiempt}x+4=5\\x+4=-5\end{array}\right.\)
<=> \(\left[\begin{array}{nghiempt}x=1\\x=-9\end{array}\right.\)
vậy x ={1;-9}