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a) Áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\frac{x}{9}=\frac{y}{8}=\frac{z}{7}=\frac{t}{6}=\frac{x-t}{9-6}=\frac{30}{3}=10\)
x/9=10 => x=90
y/8=10 => y=80
z/7=10 => z=70
t/6=10 => t=60
b) 3y=5z \(\Rightarrow\frac{y}{5}=\frac{z}{3}\)
x/4=y/3 ; y/5=z/3 \(\Rightarrow\frac{x}{20}=\frac{y}{15}=\frac{z}{9}\)
Áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\frac{x}{20}=\frac{y}{15}=\frac{z}{9}=\frac{x-y-z}{20-15-9}=\frac{100}{-4}=-25\)
x/20=-25 => x=-500
y/15=-25 => y=-375
z/9=-25 => z=-225
a)
+ Áp dụng tính chất dãy tỉ số bằng nhau, ta có
\(\frac{x}{9}=\frac{t}{6}\)⇒ \(\frac{x-t}{9-6}=\frac{30}{3}=10\)
+ Ta có:
\(\frac{x}{9}=10\)⇒x=10.9=90
\(\frac{y}{8}=10\)⇒y=10.8=80
\(\frac{z}{7}=10\)⇒z=10.7=70
\(\frac{t}{6}=10\)⇒t=10.6=60
Vậy x=90; y=80; z=70 và t=60.
Bài 1:
\(A=\frac{a+b}{b+c}.\)
Ta có:
\(\frac{b}{a}=2\Rightarrow\frac{b}{2}=\frac{a}{1}\) (1)
\(\frac{c}{b}=3\Rightarrow\frac{c}{3}=\frac{b}{1}\) (2)
Từ (1) và (2) \(\Rightarrow\frac{b}{2}=\frac{c}{6}.\)
\(\Rightarrow\frac{a}{1}=\frac{b}{2}=\frac{c}{6}=\frac{a+b}{3}=\frac{b+c}{8}.\)
\(\Rightarrow A=\frac{a+b}{b+c}=\frac{3}{8}\)
Vậy \(A=\frac{a+b}{b+c}=\frac{3}{8}.\)
Bài 2:
a) \(\frac{72-x}{7}=\frac{x-40}{9}\)
\(\Rightarrow\left(72-x\right).9=\left(x-40\right).7\)
\(\Rightarrow648-9x=7x-280\)
\(\Rightarrow648+280=7x+9x\)
\(\Rightarrow928=16x\)
\(\Rightarrow x=928:16\)
\(\Rightarrow x=58\)
Vậy \(x=58.\)
b) \(\frac{x+4}{20}=\frac{5}{x+4}\)
\(\Rightarrow\left(x+4\right).\left(x+4\right)=5.20\)
\(\Rightarrow\left(x+4\right).\left(x+4\right)=100\)
\(\Rightarrow\left(x+4\right)^2=100\)
\(\Rightarrow x+4=\pm10.\)
\(\Rightarrow\left[{}\begin{matrix}x+4=10\\x+4=-10\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=10-4\\x=\left(-10\right)-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=-14\end{matrix}\right.\)
Vậy \(x\in\left\{6;-14\right\}.\)
Chúc bạn học tốt!
Bài 2:
a, \(\frac{72-x}{7}=\frac{x-40}{9}\)
\(\Rightarrow\left(72-x\right).9=\left(x-40\right).7\)
\(\Rightarrow9.72-9.x=7.x-7.40\)
\(\Rightarrow648-9x=7x-280\)
\(\Rightarrow-9x-7x=-280-648\)
\(\Rightarrow-16x=-648\)
\(\Rightarrow x=58\)
Vậy \(x=58\)
Bài 1 :
a/ \(x^2-7x+6=0\)
\(\Leftrightarrow x^2-6x-x+6=0\)
\(\Leftrightarrow x\left(x-6\right)-\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=6\\x=1\end{matrix}\right.\)
Vậy....
b/ \(x^2-10x+9=0\)
\(\Leftrightarrow x^2-9x-x+9=0\)
\(\Leftrightarrow x\left(x-9\right)-\left(x-9\right)=0\)
\(\Leftrightarrow\left(x-9\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-9=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=1\end{matrix}\right.\)
Vậy...
c/ \(x^2+9x+8=0\)
\(\Leftrightarrow x^2+8x+x+8=0\)
\(\Leftrightarrow\left(x+8\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+8=0\\x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=-1\end{matrix}\right.\)
Vậy ...
d/ \(x^2-11x+10=0\)
\(\Leftrightarrow x^2-11x+10=0\)
\(\Leftrightarrow x^2-x-10x+10=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-10=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=10\end{matrix}\right.\)
Vậy...
Bài 2 :
Ta có :
\(\frac{2x-y}{x+y}=\frac{2}{3}\)
\(\Leftrightarrow3\left(2x-y\right)=2\left(x+y\right)\)
\(\Leftrightarrow6x-3y=2x+2y\)
\(\Leftrightarrow6x-2x=2y+3y\)
\(\Leftrightarrow4x=5y\)
\(\Leftrightarrow\frac{x}{y}=\frac{5}{4}\)
Vậy....
Bài 3 : không hiểu đề lắm ???!!!!
Bài 4 :
Ta có :
\(\frac{x}{y^2}=2\Leftrightarrow x=2y^2\left(1\right)\)
Thay (1) ta có :
\(\frac{x}{y}=16\)
\(\Leftrightarrow\frac{2y^2}{y}=16\)
\(\Leftrightarrow2y=16\)
\(\Leftrightarrow y=8\Leftrightarrow x=128\)
Vậy...
Bài 1:
a)Ta có:
\(\frac{4}{5}\left(\frac{7}{2}+\frac{1}{4}\right)^2=\frac{4}{5}\left(\frac{15}{4}\right)^2=\frac{4}{5}.\frac{15}{4}.\frac{15}{4}=\frac{45}{4}\)
b)Ta có:
\(\frac{5^4.20^4}{25^5.4^5}=\frac{\left(5.20\right)^4}{\left(25.4\right)^5}=\frac{100^4}{100^5}=\frac{1}{100}\)
Bài 2:
Ta có:
\(\frac{x}{2}=\frac{y}{-5}=\frac{x-y}{2-\left(-5\right)}=\frac{10}{7}\)
\(\Rightarrow\left\{{}\begin{matrix}x=\frac{20}{7}\\y=\frac{-50}{7}\end{matrix}\right.\)
a) Ta có: \(\left|x+\frac{3}{4}\right|+\left|y-\frac{1}{5}\right|+\left|x+y+z\right|\ge0\)
Mà \(\left|x+\frac{3}{4}\right|+\left|y-\frac{1}{5}\right|+\left|x+y+z\right|=0\)
\(\Rightarrow\left[\begin{matrix}\left|x+\frac{3}{4}\right|=0\\\left|x-\frac{1}{5}\right|=0\\\left|x+y+z\right|=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}x+\frac{3}{4}=0\\y-\frac{1}{5}=0\\x+y+z=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=\frac{-3}{4}\\y=\frac{1}{5}\\z=0-\frac{-3}{4}-\frac{1}{5}=\frac{11}{20}\end{matrix}\right.\)
Vậy \(x=\frac{-3}{4};y=\frac{1}{5};z=\frac{11}{20}\)
b) \(\left|x+\frac{3}{4}\right|+\left|y-\frac{2}{3}\right|+\left|z-\frac{1}{2}\right|=0\)
\(\Rightarrow\left[\begin{matrix}\left|x+\frac{3}{4}\right|=0\\\left|y-\frac{2}{3}\right|=0\\z+\frac{1}{2}=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}x+\frac{3}{4}=0\\y-\frac{2}{3}=0\\z+\frac{1}{2}=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=\frac{-3}{4}\\y=\frac{2}{3}\\z=\frac{-1}{2}\end{matrix}\right.\)
Vậy \(x=\frac{-3}{4};y=\frac{2}{3};z=\frac{-1}{2}\)
d) \(\left|x+1\right|+\left|x^2-1\right|=0\)
\(\Rightarrow\left[\begin{matrix}\left|x+1\right|=0\\\left|x^2-1\right|=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}x+1=0\\x^2-1=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=-1\\x=\pm1\end{matrix}\right.\)
Vậy \(x\in\left\{1;-1\right\}\)
HISINOMA KINIMADO lớp 9 bây giờ tụi anh mới được học phần nguyên :v
Bài 3:
\(\frac{a+b}{b+c}=\frac{c+d}{d+a}\Leftrightarrow\left(a+b\right)\left(d+a\right)=\left(c+d\right)\left(b+c\right)\)
\(\Leftrightarrow ad+a^2+bd+ab=bc+c^2+bd+dc\)
\(\Leftrightarrow ad+a^2+ab-bc-c^2-dc=0\)
\(\Leftrightarrow d\left(a-c\right)+b\left(a-c\right)+\left(a-c\right)\left(a+c\right)=0\)
\(\Leftrightarrow\left(a-c\right)\left(a+b+c+d\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=c\\a+b+c+d=0\end{matrix}\right.\)( đpcm )
a)
Ta có : \(\left|x+\frac{19}{5}\right|\ge0\) với mọi x
\(\left|y+\frac{1890}{1975}\right|\ge0\) với mọi x
\(\left|z-2014\right|\ge0\) với mọi x
\(\Rightarrow\left|x+\frac{19}{5}\right|+\left|y+\frac{1890}{1975}\right|+\left|z-2014\right|\ge0\)
Mà \(\left|x+\frac{19}{5}\right|+\left|y+\frac{1890}{1975}\right|+\left|z-2014\right|=0\)
\(\Rightarrow\hept{\begin{cases}\left|x+\frac{19}{5}\right|=0\\\left|y+\frac{1890}{1975}\right|=0\\\left|z-2014\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}x+\frac{19}{5}=0\\y+\frac{1890}{1975}=0\\z-2014=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-\frac{19}{5}\\y=-\frac{1890}{1975}\\z=2014\end{cases}}\)
b) Cx tương tự câu trên thôi bạn
Ta có : \(\left|x-\frac{9}{2}\right|\ge0\) với mọi x
\(\left|y+\frac{4}{3}\right|\ge0\) với mọi x
\(\left|z+\frac{7}{2}\right|\ge0\) với mọi x
\(\Rightarrow\left|x-\frac{9}{2}\right|+\left|y+\frac{4}{3}\right|+\left|z+\frac{7}{2}\right|\ge0\) với mọi x
Mà \(\left|x-\frac{9}{2}\right|+\left|y+\frac{4}{3}\right|+\left|z+\frac{7}{2}\right|\le0\)
\(\Rightarrow\hept{\begin{cases}\left|x-\frac{9}{2}\right|=0\\\left|y+\frac{4}{3}\right|=0\\\left|z+\frac{7}{2}\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}x-\frac{9}{2}=0\\y+\frac{4}{3}=0\\z+\frac{7}{2}=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{9}{2}\\y=-\frac{4}{3}\\z=-\frac{7}{2}\end{cases}}\)