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Đặt \(\frac{3\left|x\right|+5}{3}=\frac{3\left|y\right|-1}{5}=\frac{3-z}{7}=k\)
\(\Rightarrow\left|x\right|=\frac{3k-5}{3}\Rightarrow2\left|x\right|=\frac{6k-10}{3}\)
\(\Rightarrow\left|y\right|=\frac{5k+1}{3}\Rightarrow7\left|y\right|=\frac{35k+7}{3}\)
\(\Rightarrow z=3-7k\Rightarrow3z=9-21k\)
Vì \(2\left|x\right|+7\left|y\right|+3z=-14\)\(\Rightarrow\frac{6k-10}{3}+\frac{35k+7}{3}+\left(9-21k\right)=-14\)
\(\Rightarrow\frac{\left(6k-10\right)+\left(35k+7\right)+\left(27-63k\right)}{3}=-14\)
\(\Rightarrow\frac{-22k+24}{3}=-14\)
\(\Rightarrow-22k+24=-42\)
\(\Rightarrow k=\frac{-42-24}{22}=3\)
\(\Rightarrow\left|x\right|=\frac{3.3-5}{3}=\frac{4}{3}\Rightarrow x=-\frac{4}{3};\frac{4}{3}\)
\(\Rightarrow\left|y\right|=\frac{5.3+1}{3}=\frac{16}{3}\Rightarrow y=-\frac{16}{3};\frac{16}{3}\)
\(\Rightarrow z=3-7.3=-18\)
\(a,\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)
\(\Rightarrow\left(x-1\right)^{x+2}\left[1-\left(x-1\right)^2\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-1\right)^{x+2}=0\\1-\left(x-1\right)^2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x\left(2-x\right)=0\end{cases}}}\)
=> x=1 ; x=0 ; x=2
Vậy..
Bài 1 :
b) \(\left|x-3\right|=5\)
\(\Rightarrow\orbr{\begin{cases}x-3=-5\\x-3=5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-2\\x=8\end{cases}}\)
Vậy x thuộc {-2; 8}
c) \(\left|2x+1\right|=x-8\)
\(\Rightarrow\orbr{\begin{cases}2x+1=-x+8\\2x+1=x-8\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x=7\\x=-9\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x=-9\end{cases}}\)
Vậy x thuộc {-9; 7/3}
Câu c) tớ không chắc, thông cảm.
=))
a) \(\left|2x+\frac{3}{4}\right|=\frac{1}{2}\)
\(\orbr{\begin{cases}2x+\frac{3}{4}=\frac{1}{2}\\2x+\frac{3}{4}=\frac{-1}{2}\end{cases}}\) => \(\orbr{\begin{cases}2x=\frac{1}{2}-\frac{3}{4}\\2x=\frac{-1}{2}-\frac{3}{4}\end{cases}}\) => \(\orbr{\begin{cases}2x=\frac{-1}{4}\\2x=\frac{-5}{4}\end{cases}}\) => \(\orbr{\begin{cases}x=\frac{-1}{8}\\x=\frac{-5}{8}\end{cases}}\)
Vậy \(x=\left\{\frac{-1}{8},\frac{-5}{8}\right\}\)
b) \(\frac{3x}{2,7}=\frac{\frac{1}{4}}{2\frac{1}{4}}\)= \(\frac{3x}{2,7}=\frac{\frac{1}{4}}{\frac{9}{4}}\)
=> \(3x.\frac{9}{4}=2,7.\frac{1}{4}\)=> \(\frac{27x}{4}=\frac{27}{40}\)
\(27x.40=27.4\)
\(1080.x=108\)
\(x=\frac{1}{10}\)
Vậy \(x=\frac{1}{10}\)
c) \(\left|x-1\right|+4=6\)
\(\left|x-1\right|=6-4\)
\(\left|x-1\right|=2\)
\(\orbr{\begin{cases}x-1=2\\x-1=-2\end{cases}}\)=> \(\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
Vậy \(x=\left[3,-1\right]\)
d) \(\frac{x}{3}=\frac{y}{5}=>\frac{y}{5}=\frac{x}{3}=>\frac{y-x}{5-3}=\frac{24}{2}=12\)
e) \(\left(x^2-3\right)^2=16\)
\(\left(x^2-3\right)^2=4^2\)\(=>x^2-3=4\)
\(x^2=7=>x=\sqrt{7}\)
Vậy \(x=\sqrt{7}\)
f) \(\frac{3}{4}+\frac{2}{5}x=\frac{29}{60}\)
\(\frac{2}{5}x=\frac{29}{60}-\frac{3}{4}\)
\(\frac{2}{5}x=-\frac{4}{15}\)
\(x=-\frac{4}{15}:\frac{2}{5}=-\frac{4}{15}.\frac{5}{2}=-\frac{2}{3}\)
Vậy \(x=-\frac{2}{3}\)
g) \(\left(-\frac{1}{3}\right)^3.x=\frac{1}{81}\)
\(\left(-\frac{1}{27}\right).x=\frac{1}{81}\)
\(x=\left(-\frac{1}{27}\right):\frac{1}{81}=\left(-\frac{1}{27}\right).81=-3\)
Vậy \(x=-3\)
k)\(\frac{3}{4}-\frac{2}{5}x=\frac{29}{60}\)
\(\frac{2}{5}x=\frac{3}{4}-\frac{29}{60}\)
\(\frac{2}{5}x=\frac{4}{15}\)
\(x=\frac{2}{5}-\frac{4}{15}=>x=\frac{2}{15}\)
Vậy \(x=\frac{2}{15}\)
I) \(\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)
\(\frac{3}{5}x=-\frac{1}{7}+\frac{1}{2}\)
\(\frac{3}{5}x=\frac{5}{14}\)
\(x=\frac{5}{14}:\frac{3}{5}=\frac{5}{14}.\frac{5}{3}=\frac{25}{42}\)
Vậy \(x=\frac{25}{42}\)
a) Ta có: \(\frac{x}{12}=\frac{y}{3}.\)
=> \(\frac{x}{12}=\frac{y}{3}\) và \(x-y=36.\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\frac{x}{12}=\frac{y}{3}=\frac{x-y}{12-3}=\frac{36}{9}=4.\)
\(\left\{{}\begin{matrix}\frac{x}{12}=4=>x=4.12=48\\\frac{y}{3}=4=>y=4.3=12\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(48;12\right).\)
b)
\(\frac{2}{3}+\frac{5}{3}x=\frac{5}{7}\)
⇒ \(\frac{5}{3}x=\frac{5}{7}-\frac{2}{3}\)
⇒ \(\frac{5}{3}x=\frac{1}{21}\)
⇒ \(x=\frac{1}{21}:\frac{5}{3}\)
⇒ \(x=\frac{1}{35}\)
Vậy \(x=\frac{1}{35}.\)
\(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
⇒ \(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
⇒ \(x-\frac{1}{2}=\frac{1}{3}\)
⇒ \(x=\frac{1}{3}+\frac{1}{2}\)
⇒ \(x=\frac{5}{6}\)
Vậy \(x=\frac{5}{6}.\)
Có 1 câu bạn đăng mình làm ở dưới rồi mà.
Chúc bạn học tốt!
a)áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{12}=\frac{y}{3}=\frac{x-y}{12-3}=\frac{36}{9}=4\)
\(\)x/12=4 suy ra x=12.4=48
y/3=4 suy ra y=3.4 =12
b)\(\frac{2}{3}+\frac{5}{3}x=\frac{5}{7}\)
\(\frac{5}{3}x=\frac{5}{7}-\frac{2}{3}\)
\(\frac{5}{3}x=\frac{1}{21}\)
\(x=\frac{1}{21}:\frac{5}{3}\)
\(x=\frac{1}{35}\)
\(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{3}\)
\(\left(\frac{2}{5}+x\right)=\frac{11}{12}-\frac{2}{3}\)
\(\frac{2}{5}+x=\frac{1}{4}\)
\(x=\frac{1}{4}-\frac{2}{5}\)
\(x=\frac{-3}{20}\)
\(\left|x-\frac{2}{5}\right|+\frac{3}{4}=\frac{11}{4}\)
\(\left|x-\frac{2}{5}\right|=\frac{11}{4}-\frac{3}{4}\)
\(\left|x-\frac{2}{5}\right|=2\)
suy ra x-2/5=2 hoac x-2/5=-2
\(x-\frac{2}{5}=2\)
\(x=\frac{12}{5}\)
\(x-\frac{2}{5}=-2\)
\(x=\frac{-8}{5}\)
\(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
\(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
\(x-\frac{1}{2}=\frac{1}{3}\)
\(x=\frac{1}{3}+\frac{1}{2}\)
\(x=\frac{5}{6}\)
a, \(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{5x}{15}=\frac{2y}{8}=\frac{5x-2y}{15-8}=\frac{28}{7}=4\)
=> x = 4.3 = 12
y = 4.4 = 16
b, \(x:2=y:\left(-5\right)\Rightarrow\frac{x}{2}=\frac{y}{-5}=\frac{x-y}{2-\left(-5\right)}=\frac{-7}{7}=-1\)
=> x = (-1).2 = -2
y = (-1)(-5) = 5
c, \(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{8}=\frac{y}{12}\)
\(\frac{y}{4}=\frac{z}{5}\Rightarrow\frac{y}{12}=\frac{z}{15}\)
\(\Rightarrow\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{x+y-z}{8+12-10}=\frac{10}{10}=1\)
=> x = 8
y =12
z = 15
bày em cách làm với được không ạ? em tự suy ra chứ thầy cô chưa bày j cả nên là em cx chưa hiểu cho lắm mong anh giúp đỡ ạ
a) \(A=x\cdot\left(-1\right)^n\cdot\left|x\right|\)
\(A=x\cdot\left(-1\right)\cdot x\)
\(A=-x^2\)
b) \(\frac{x}{y}-\frac{2}{3}=\frac{y}{z}-\frac{4}{5}=\frac{z}{t}-\frac{6}{7}=0\)và \(x+y+z+t=315\)
Xét :
\(\frac{x}{y}-\frac{2}{3}=0\Leftrightarrow\frac{x}{y}=\frac{2}{3}\Leftrightarrow\frac{x}{2}=\frac{y}{3}\Leftrightarrow\frac{x}{8}=\frac{y}{12}\)
\(\frac{y}{z}-\frac{4}{5}=0\Leftrightarrow\frac{y}{z}=\frac{4}{5}\Leftrightarrow\frac{y}{4}=\frac{z}{5}\Leftrightarrow\frac{y}{12}=\frac{z}{15}\)
\(\frac{z}{t}-\frac{6}{7}=0\Leftrightarrow\frac{z}{t}=\frac{6}{7}\Leftrightarrow\frac{z}{6}=\frac{t}{7}\Leftrightarrow\frac{z}{15}=\frac{t}{\frac{35}{2}}\)
\(\Rightarrow\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{t}{\frac{35}{2}}\) và \(x+y+z+t=315\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{t}{\frac{35}{2}}=\frac{x+y+z+t}{8+12+15+\frac{35}{2}}=\frac{315}{\frac{105}{2}}=6\)
\(\frac{x}{8}=6\Leftrightarrow x=48\)
\(\frac{y}{12}=6\Leftrightarrow y=72\)
\(\frac{z}{15}=6\Leftrightarrow z=90\)
\(\frac{t}{\frac{35}{2}}=6\Leftrightarrow t=105\)
ta có
\(\frac{x}{y}-\frac{2}{3}=0\Leftrightarrow\frac{x}{y}=\frac{2}{3}\Leftrightarrow\frac{x}{2}=\frac{y}{3}\)
\(\frac{y}{z}-\frac{4}{5}=0\Leftrightarrow\frac{y}{z}=\frac{4}{5}\Leftrightarrow\frac{y}{4}=\frac{z}{5}\)
\(\frac{z}{t}-\frac{6}{7}=0\Leftrightarrow\frac{z}{t}=\frac{6}{7}\Leftrightarrow\frac{z}{7}=\frac{t}{6}\)
ta lại có
\(\hept{\begin{cases}\frac{x}{2}=\frac{y}{3}\\\frac{y}{4}=\frac{z}{5}\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{x}{8}=\frac{y}{12}\\\frac{y}{12}=\frac{z}{15}\end{cases}}}\Leftrightarrow\frac{x}{8}=\frac{y}{12}=\frac{z}{15}\left(1\right)\)
\(\hept{\begin{cases}\frac{y}{12}=\frac{z}{15}\\\frac{z}{7}=\frac{t}{6}\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{y}{84}=\frac{z}{105}\\\frac{z}{105}=\frac{t}{90}\end{cases}}}\Leftrightarrow\frac{y}{84}=\frac{z}{105}=\frac{t}{90}\left(2\right)\)
ta kết hợp (1) và (2)
\(\hept{\begin{cases}\frac{x}{8}=\frac{y}{12}=\frac{z}{15}\\\frac{y}{84}=\frac{z}{105}=\frac{t}{90}\end{cases}}\Leftrightarrow\frac{x}{57}=\frac{y}{84}=\frac{z}{105}=\frac{t}{90}\)và \(x+y+z+t=315\)
theo tính chất dãy tỉ số = nhau
có \(\frac{x}{57}=\frac{y}{84}=\frac{z}{105}=\frac{t}{90}=\frac{x+y+z+t}{57+84+105+90}=\frac{315}{336}=\frac{15}{16}\)
thay vào
Áp dụng tính chất dãy tỉ số = , có:
x/3 = y/4 = (x+y)/(3+4) = 5/7
Vì x/3 = 5/7 => x= 15/7
y/4 = 5/7 => y = 20/7
7(x-y) = 7(15/7 - 20/7) = -5