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Bài 1:
a: \(\dfrac{-3}{5}+\dfrac{7}{21}+\dfrac{-4}{5}+\dfrac{7}{5}\)
\(=\left(-\dfrac{3}{5}-\dfrac{4}{5}+\dfrac{7}{5}\right)+\dfrac{7}{21}\)
\(=\dfrac{7}{21}=\dfrac{1}{3}\)
b: \(-\dfrac{3}{17}+\left(\dfrac{2}{3}+\dfrac{3}{17}\right)\)
\(=-\dfrac{3}{17}+\dfrac{3}{17}+\dfrac{2}{3}\)
\(=0+\dfrac{2}{3}=\dfrac{2}{3}\)
c: \(\dfrac{-5}{21}+\left(\dfrac{-16}{21}+1\right)\)
\(=\left(-\dfrac{5}{21}-\dfrac{16}{21}\right)+1\)
=-1+1=0
d: \(\dfrac{5}{7}+\dfrac{9}{23}+\dfrac{-12}{7}+\dfrac{14}{23}\)
\(=\left(\dfrac{5}{7}-\dfrac{12}{7}\right)+\left(\dfrac{9}{23}+\dfrac{14}{23}\right)\)
\(=-1+1=0\)
e: \(\dfrac{3}{17}+\dfrac{-5}{13}+\dfrac{-18}{35}+\dfrac{14}{17}+\dfrac{17}{-35}+\dfrac{-8}{13}\)
\(=\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(-\dfrac{5}{13}-\dfrac{8}{13}\right)+\left(-\dfrac{18}{35}-\dfrac{17}{35}\right)\)
=1-1-1
=-1
f: \(\dfrac{-3}{8}\cdot\dfrac{1}{6}+\dfrac{3}{-8}\cdot\dfrac{5}{6}+\dfrac{-10}{16}\)
\(=\dfrac{-3}{8}\left(\dfrac{1}{6}+\dfrac{5}{6}\right)+\dfrac{-5}{8}\)
\(=-\dfrac{3}{8}-\dfrac{5}{8}=-1\)
g: \(\dfrac{-4}{11}\cdot\dfrac{5}{15}\cdot\dfrac{11}{-4}=\dfrac{-4}{-4}\cdot\dfrac{11}{11}\cdot\dfrac{1}{3}=\dfrac{1}{3}\)
h: \(\dfrac{7}{36}-\dfrac{8}{-9}+\dfrac{-2}{3}\)
\(=\dfrac{7}{36}+\dfrac{8}{9}-\dfrac{2}{3}\)
\(=\dfrac{7}{36}+\dfrac{32}{36}-\dfrac{24}{36}=\dfrac{15}{36}=\dfrac{5}{12}\)
i: \(\dfrac{4}{7}+\dfrac{-5}{8}-\dfrac{3}{28}\)
\(=\dfrac{32}{56}-\dfrac{35}{56}-\dfrac{6}{56}\)
\(=-\dfrac{9}{56}\)
l: \(\dfrac{-6}{11}:\left(\dfrac{3}{5}\cdot\dfrac{4}{11}\right)\)
\(=-\dfrac{6}{11}:\dfrac{12}{55}\)
\(=-\dfrac{6}{11}\cdot\dfrac{55}{12}=\dfrac{-5}{2}\)
Dạng 2:
a: \(1\dfrac{3}{4}x-5=3\dfrac{1}{3}\)
=>\(x\cdot\dfrac{7}{4}-5=\dfrac{10}{3}\)
=>\(x\cdot\dfrac{7}{4}=\dfrac{10}{3}+5=\dfrac{25}{3}\)
=>\(x=\dfrac{25}{3}:\dfrac{7}{4}=\dfrac{25}{3}\cdot\dfrac{4}{7}=\dfrac{100}{21}\)
b: \(\dfrac{2}{3}x+\dfrac{1}{4}=\dfrac{7}{12}\)
=>\(\dfrac{2}{3}x=\dfrac{7}{12}-\dfrac{1}{4}=\dfrac{7}{12}-\dfrac{3}{12}=\dfrac{4}{12}=\dfrac{1}{3}\)
=>\(x=\dfrac{1}{3}:\dfrac{2}{3}=\dfrac{1}{2}\)
c: \(\dfrac{1}{3}+\dfrac{2}{5}\left(x+1\right)=1\)
=>\(\dfrac{2}{5}\left(x+1\right)=1-\dfrac{1}{3}=\dfrac{2}{3}\)
=>\(x+1=\dfrac{2}{3}:\dfrac{2}{5}=\dfrac{5}{3}\)
=>\(x=\dfrac{5}{3}-1=\dfrac{2}{3}\)
d: \(\dfrac{1}{4}+\dfrac{1}{3}:3x=-5\)
=>\(\dfrac{1}{3}:3x=-5-\dfrac{1}{4}=-\dfrac{21}{4}\)
=>\(3x=\dfrac{1}{3}:\dfrac{-21}{4}=\dfrac{1}{3}\cdot\dfrac{4}{-21}=\dfrac{-4}{63}\)
=>\(x=-\dfrac{4}{63}:3=-\dfrac{4}{189}\)
e: \(2x^2-72=0\)
=>\(2x^2=72\)
=>\(x^2=36\)
=>\(\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)
f: \(\left(\dfrac{3}{5}x-0,75\right):\dfrac{3}{7}=2\dfrac{4}{5}\)
=>\(\left(\dfrac{3}{5}x-0,75\right):\dfrac{3}{7}=\dfrac{14}{5}\)
=>\(\dfrac{3}{5}x-0,75=\dfrac{14}{5}\cdot\dfrac{3}{7}=\dfrac{6}{5}\)
=>\(\dfrac{3}{5}x=\dfrac{6}{5}+\dfrac{3}{4}=\dfrac{39}{20}\)
=>\(x=\dfrac{39}{20}:\dfrac{3}{5}=\dfrac{39}{20}\cdot\dfrac{5}{3}=\dfrac{13}{4}\)
g: \(2x+\dfrac{3}{10}=1\dfrac{5}{6}\cdot\dfrac{6}{11}\)
=>\(2x+\dfrac{3}{10}=\dfrac{11}{6}\cdot\dfrac{6}{11}=1\)
=>\(2x=\dfrac{7}{10}\)
=>\(x=\dfrac{7}{20}\)
h: \(2\dfrac{1}{4}:\left(x-7\dfrac{1}{3}\right)=-1,5\)
=>\(\dfrac{9}{4}:\left(x-\dfrac{22}{3}\right)=-\dfrac{3}{2}\)
=>\(x-\dfrac{22}{3}=\dfrac{-9}{4}:\dfrac{3}{2}=-\dfrac{9}{4}\cdot\dfrac{2}{3}=\dfrac{-3}{2}\)
=>\(x=-\dfrac{3}{2}+\dfrac{22}{3}=\dfrac{35}{6}\)
a: \(2x-\dfrac{5}{4}=\dfrac{3}{2}\)
=>\(2x=\dfrac{3}{2}+\dfrac{5}{4}=\dfrac{6}{4}+\dfrac{5}{4}=\dfrac{11}{4}\)
=>\(x=\dfrac{11}{8}\)
b: \(\dfrac{1}{5}:x-\dfrac{6}{7}=\dfrac{3}{14}\)
=>\(\dfrac{1}{5}:x=\dfrac{3}{14}+\dfrac{6}{7}=\dfrac{3}{14}+\dfrac{12}{14}=\dfrac{15}{14}\)
=>\(x=\dfrac{1}{5}:\dfrac{15}{14}=\dfrac{1}{5}\cdot\dfrac{14}{15}=\dfrac{14}{75}\)
c: \(x:\dfrac{4}{9}+\dfrac{5}{9}=\dfrac{13}{3}\)
=>\(x:\dfrac{4}{9}=\dfrac{13}{3}-\dfrac{5}{9}=\dfrac{39}{9}-\dfrac{5}{9}=\dfrac{34}{9}\)
=>\(x=\dfrac{34}{9}\cdot\dfrac{4}{9}=\dfrac{136}{81}\)
d: \(17-x\times\dfrac{8}{3}=\dfrac{1}{2}\)
=>\(x\times\dfrac{8}{3}=17-\dfrac{1}{2}=\dfrac{33}{2}\)
=>\(x=\dfrac{33}{2}:\dfrac{8}{3}=\dfrac{33}{2}\times\dfrac{3}{8}=\dfrac{99}{16}\)
e: \(\dfrac{21}{4}+x:\dfrac{5}{2}=\dfrac{3}{2}\)
=>\(x:\dfrac{5}{2}=\dfrac{3}{2}-\dfrac{21}{4}=\dfrac{-15}{4}\)
=>\(x=-\dfrac{15}{4}\times\dfrac{5}{2}=-\dfrac{75}{8}\)
g: \(\dfrac{18}{2}:2-4:x=\dfrac{3}{10}\)
=>\(4:x=\dfrac{9}{2}-\dfrac{3}{10}=\dfrac{45}{10}-\dfrac{3}{10}=\dfrac{42}{10}=\dfrac{21}{5}\)
=>\(x=4:\dfrac{21}{5}=\dfrac{20}{21}\)
a; 2\(x\) - \(\dfrac{5}{4}\) = \(\dfrac{3}{2}\)
2\(x\) = \(\dfrac{3}{2}\) + \(\dfrac{5}{4}\)
2\(x\) = \(\dfrac{11}{4}\)
\(x\) = \(\dfrac{11}{4}\) : 2
\(x\) = \(\dfrac{11}{8}\)
Vì AD//BC
nên \(\dfrac{OA}{OC}=\dfrac{OD}{OB}=k\)
=>\(OC=k\times OA;OB=k\times OD\)
Vì \(OC=k\times OA\)
nên \(S_{DOC}=k\times S_{AOD}\)
Vì \(OB=k\times OD\)
nên \(S_{AOB}=k\times S_{AOD}\)
Do đó: \(S_{AOB}=S_{DOC}\)
Vì AD/BC
Nên OA/OC=OD/OB=k
=>OC=kxOA;OB=kxOD
Vì OC=kxOA
Nên sDOC=kx sAOD
Vì OB=kxOD
Nên sAOB=kx sAOD
Do đó:sAOB=sDOC
Tick cho mình nhé
\(\left\{{}\begin{matrix}8x-y=6\\x^2-y=-6\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}8x-y-x^2+y=6+6\\8x-y=6\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x^2-8x=-12\\y=8x-6\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x^2-8x+12=0\\y=8x-6\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(x-2\right)\left(x-6\right)=0\\y=8x-6\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\in\left\{2;6\right\}\\y=8x-6\end{matrix}\right.\)
Khi x=2 thì \(y=8\cdot2-6=16-6=10\)
Khi x=6 thì \(y=8\cdot6-6=42\)
Câu 26:
a: Vì D là trung điểm của BC
nên \(S_{ADB}=S_{ADC}=\dfrac{S_{ABC}}{2}\)
Vì \(AM=\dfrac{1}{3}AC\)
nên \(S_{AMD}=\dfrac{1}{3}\times S_{ADC}=\dfrac{1}{3}\times\dfrac{1}{2}\times S_{ABC}=\dfrac{1}{6}\times S_{ABC}\)
b: Vì \(AM=\dfrac{1}{3}AC\)
nên \(S_{ABM}=\dfrac{1}{3}\times S_{ABC}\)
Vì AN=NB
nên N là trung điểm của AB
=>\(AN=\dfrac{1}{2}AB=NB\)
\(AN=\dfrac{1}{2}AB\)
=>\(S_{ANM}=\dfrac{1}{2}\times S_{ABM}=\dfrac{1}{6}\times S_{ABC}\)
Vì \(BM=\dfrac{1}{2}AB\)
nên \(S_{BND}=\dfrac{1}{2}\times S_{ABD}=\dfrac{1}{4}\times S_{ABC}\)
Vì \(AM=\dfrac{1}{3}AC\)
nên \(CM=\dfrac{2}{3}AC\)
=>\(S_{DMC}=\dfrac{2}{3}\times S_{ADC}=\dfrac{1}{3}\times S_{ABC}\)
Ta có: \(S_{ANM}+S_{DNM}+S_{BND}+S_{MDC}=S_{ABC}\)
=>\(S_{DNM}+\dfrac{1}{3}\times S_{ABC}+\dfrac{1}{4}\times S_{ABC}+\dfrac{1}{6}\times S_{ABC}=S_{ABC}\)
=>\(S_{DNM}=\dfrac{1}{4}\times S_{ABC}=150\left(cm^2\right)\)
Câu 25:
Tỉ số giữa số bi xanh và số bi đỏ là:
\(\dfrac{1}{8}:\dfrac{1}{9}=\dfrac{9}{8}\)
Số bi xanh là: 170:17x9=90(viên)
Số bi đỏ là 170-90=80(viên)
Ctv vip là một cộng tác viên vip của Olm so với ctv thì ctv vip có thể tick ra gp em nhé.
Thông thường thì hai nghiệm phải có quan hệ với nhau, sao biểu thức trong căn chỉ chứa \(x_1\) vậy em?
Bài nào em chưa biết cách làm thì hỏi để Thầy cô và các bạn hướng dẫn. Em không gửi một tệp bài lên nhờ mọi người giải như vậy em sẽ không học tập phát triển được.
Dạ vâng ạ!