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1.b) \(\left(\left|x\right|-3\right)\left(x^2+4\right)< 0\)
\(\Rightarrow\hept{\begin{cases}\left|x\right|-3\\x^2+4\end{cases}}\) trái dấu
\(TH1:\hept{\begin{cases}\left|x\right|-3< 0\\x^2+4>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|< 3\\x^2>-4\end{cases}}\Leftrightarrow x\in\left\{0;\pm1;\pm2\right\}\)
\(TH1:\hept{\begin{cases}\left|x\right|-3>0\\x^2+4< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|>3\\x^2< -4\end{cases}}\Leftrightarrow x\in\left\{\varnothing\right\}\)
Vậy \(x\in\left\{0;\pm1;\pm2\right\}\)
Bài 1: Làm:
a,
- x - 2/3 = - 6/7
<=> - x = - 6/7 + 2/3 = -18/21 + 14/21
<=> - x = - 4/21
<=> x = 4/21.
Vậy x = 4/21.
b,
x/- 27 = - 3 / x
<=> x^2 = - 27 . (- 3)
<=> x^2 = 81
<=> x thuộc {9;- 9}
Vậy x thuộc {9;- 9}.
c,
x / y = 2 / 5
<=> x / 2 = y / 5 = 2x - y / 2.2 - 5 = 3 / -1 = - 3.
(T/c dãy tỷ số bằng nhau)
=> x / 2 = - 3 <=> x = - 6.
y / 5 = - 3 <=> y = - 15.
Vậy x = - 6 ; y = - 15.
Bài 2: Làm:
1/2 a = 2/3 b = 3/4 c
<=> a/2 = 2b/3 = 3c/4
<=> a/2.6 = 2b/3.6 = 3c/4.6 (mỗi vế nhân với 1/6)
<=> a/12 = 2b/18 = 3c/24
<=> a/12 = b/9 = c/8 (Rút gọn)
Áp dụng tính chất dãy tỷ số bằng nhau ta có:
a/12 = b/9 = c/8 = a - b/ 12 - 9 = 15 / 3 = 5 (Theo đề bài)
=> a/12 = 3 <=>a = 36
b/9 = 3 <=> b = 27
c/8 = 3 <=> c = 24
Vậy a = 36 ; b = 27 ; c = 24.
Học tốt !
.a, \(\frac{x+1}{999}+\frac{x+2}{998}=\frac{x+3}{997}+\frac{x+4}{996}\)
.\(< =>\frac{x+1}{999}+1+\frac{x+2}{998}+1=\frac{x+3}{997}+1+\frac{x+4}{996}+1\)
.\(< =>\frac{x+1}{999}+\frac{999}{999}+\frac{x+2}{998}+\frac{998}{998}=\frac{x+3}{997}+\frac{997}{997}+\frac{x+4}{996}+\frac{996}{996}\)
.\(< =>\frac{x+1+999}{999}+\frac{x+2+998}{998}=\frac{x+3+997}{997}+\frac{x+4+996}{996}\)
.\(< =>\frac{x+1000}{999}+\frac{x+1000}{998}-\frac{x+1000}{997}-\frac{x+1000}{996}=0\)
.\(< =>\left(x+1000\right)\left(\frac{1}{999}+\frac{1}{998}-\frac{1}{997}-\frac{1}{996}\right)=0\)
.Do \(\frac{1}{999}+\frac{1}{998}-\frac{1}{997}-\frac{1}{996}\ne0\)
.Suy ra \(x+1000=0\Leftrightarrow x=-1000\)
.b, \(\frac{x+1}{1001}+\frac{x+2}{1002}=\frac{x+3}{1003}+\frac{x+4}{1004}\)
.\(< =>\frac{x+1}{1001}-1+\frac{x+2}{1002}-1=\frac{x+3}{1003}-1+\frac{x+4}{1004}-1\)
.\(< =>\frac{x+1}{1001}-\frac{1001}{1001}+\frac{x+2}{1002}-\frac{1002}{1002}=\frac{x+3}{1003}-\frac{1003}{1003}+\frac{x+4}{1004}-\frac{1004}{1004}\)
.\(< =>\frac{x+1-1001}{1001}+\frac{x+2-1002}{1002}=\frac{x+3-1003}{1003}+\frac{x+4-1004}{1004}\)
.\(< =>\frac{x-1000}{1001}+\frac{x+1000}{1002}-\frac{x+1000}{1003}-\frac{x+1000}{1004}=0\)
.\(< =>\left(x-1000\right)\left(\frac{1}{1001}+\frac{1}{1002}-\frac{1}{1003}-\frac{1}{1004}\right)=0\)
.Do \(\frac{1}{1001}+\frac{1}{1002}-\frac{1}{1003}-\frac{1}{1004}\ne0\)
.Suy ra \(x-1000=0\Leftrightarrow x=1000\)
Bài 1
\(=-\frac{21}{60}=-\frac{7}{20}\)
\(b,\left(2-\frac{1}{3}\right)^2+|-\frac{5}{6}|+\frac{-7}{12}-\frac{25}{9}\)
\(=\frac{25}{9}+\frac{5}{6}-\frac{7}{12}-\frac{25}{9}\)
\(=\left(\frac{25}{9}-\frac{25}{9}\right)+\left(\frac{5}{6}-\frac{7}{12}\right)\)
\(=0+\frac{1}{4}=\frac{1}{4}\)
Bài 2
\(a,x+\frac{2}{5}=-\frac{3}{10}\)
\(x=-\frac{3}{10}-\frac{2}{5}\)
\(x=-\frac{3}{10}-\frac{4}{10}\)
\(x=-\frac{7}{10}\)
\(b,|\frac{2}{3}+x|=\frac{5}{7}\)
\(\Rightarrow\orbr{\begin{cases}\frac{2}{3}+x=\frac{5}{7}\\\frac{2}{3}+x=-\frac{5}{7}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{5}{7}-\frac{2}{3}\\x=-\frac{5}{7}-\frac{2}{3}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{21}\\x=-\frac{29}{21}\end{cases}}}\)
== chắc trog quá trình lm lỡ xóa đó
\(a,-\frac{3}{4}.\frac{7}{15}\)
\(=-\frac{21}{60}=-\frac{7}{20}\)
với lại bài trên mk tính nhẩm ko bấm máy sai == sửa giúp
1a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\)
=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)
b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c) TT
a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\-\frac{3}{2}x-\frac{1}{2}=4x-1\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}-4x=-1\\-\frac{3}{2}x-\frac{1}{2}-4x=-1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)
\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)
=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)
=> \(\left|50x-140\right|=\left|25x+24\right|\)
=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)
=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)
Bài 2 : a. |2x - 5| = x + 1
TH1 : 2x - 5 = x + 1
=> 2x - 5 - x = 1
=> 2x - x - 5 = 1
=> 2x - x = 6
=> x = 6
TH2 : -2x + 5 = x + 1
=> -2x + 5 - x = 1
=> -2x - x + 5 = 1
=> -3x = -4
=> x = 4/3
Ba bài còn lại tương tự
2a) \(\frac{3^6+45^4-15^3.4^5}{27^4.25^3+45^6}\)
= \(\frac{3^6+\left(3^2.5\right)^4-\left(3.5\right)^3.\left(2^2\right)^5}{\left(3^3\right)^4.\left(5^2\right)^3+\left(3^2.5\right)^6}\)
= \(\frac{3^6+3^8.5^4-3^3.5^3.4^{10}}{3^{12}.5^6-3^{12}.5^6}=\frac{3^3.\left(3^3+3^5.5^4-5^3.4^{10}\right)}{0}\)(xem lại đề)
b) \(\frac{\left(\frac{2}{5}\right)^7.5^7+\left(\frac{16}{3}\right)^3:\left(\frac{4}{9}\right)^3}{2^7.5^2+512}\)
= \(\frac{\left(\frac{2}{5}.5\right)^7+\left(\frac{16}{3}:\frac{4}{9}\right)^3}{2^7.5^2+2^9}\)
= \(\frac{2^7+12^3}{2^7\left(5^2+2^2\right)}\)
= \(\frac{2^7+\left(2^2.3\right)^3}{2^7.29}\)
= \(\frac{2^7+2^6.3^3}{2^7.29}\)
= \(\frac{2^6\left(1+27\right)}{2^7.29}=\frac{28}{2.29}=\frac{14}{29}\)
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)15-3 x (x+2)=-21
3 x (x +2)=15-(-21)=36
x+2=36/3=12
x=12-2
x=10
b)\(\left(x-2\right).2=\left(x+3\right).3\)
\(\left(x+1\right).2=3\)
\(x+1=6\)
\(x=6-1=5\)
a) 15-3(x+2)=-21
3(x+2)=15-(-21)=36
x+2=12
x=10
b) \(\frac{x-2}{3}=\frac{x+3}{2}\)
=>2(x-2)=3(x+3)
=>2x-4=3x+9
=>3x-2x=-4-9
=>x=-13