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\(\frac{3x+25}{144}=\frac{2y-169}{25}=\frac{z+144}{169}=\frac{3x+2y+z}{338}=\frac{169}{338}=\frac{1}{2}\)
\(\Rightarrow3x+25=\frac{1}{2}.144=72\)
\(\Leftrightarrow x=\frac{47}{3}\)
\(2y-169=\frac{1}{2}.25=\frac{25}{2}\)
\(\Leftrightarrow y=\frac{363}{4}\)
\(z+144=\frac{1}{2}.169=\frac{169}{2}\)
\(\Leftrightarrow z=\frac{-119}{2}\)
Áp dụng tính chất dãy tỉ số bằng nhau
\(\frac{3x+25}{144}=\frac{2y-169}{25}=\frac{z+144}{169}=\frac{\left(3x+2y+z\right)+\left(25-169+144\right)}{144+25+169}=\frac{169+25-169+144}{144+25+169}=\)
\(\frac{1}{2}\)
Ta có
\(\frac{3x+25}{144}=\frac{1}{2}\Rightarrow6x+50=144\Rightarrow6x=94\Rightarrow x=\frac{47}{3}\)
\(\frac{2y-169}{25}=\frac{1}{2}\Rightarrow4y-338=25\Rightarrow4y=363\Rightarrow y=\frac{363}{4}\)
\(\frac{z+144}{169}=\frac{1}{2}\Rightarrow2z+288=169\Rightarrow2z=-119\Rightarrow z=\frac{-119}{2}\)
\(\dfrac{148-x}{25}+\dfrac{169-x}{23}+\dfrac{186-x}{21}+\dfrac{199-x}{19}=10\)
\(\Leftrightarrow\left(\dfrac{148-x}{25}-1\right)+\left(\dfrac{169-x}{23}-2\right)+\left(\dfrac{186-x}{21}-3\right)+\left(\dfrac{199-x}{19}-4\right)=0\)
\(\Leftrightarrow\dfrac{123-x}{25}+\dfrac{123-x}{23}+\dfrac{123-x}{21}+\dfrac{123-x}{19}=0\)
\(\Leftrightarrow\left(123-x\right)\left(\dfrac{1}{25}+\dfrac{1}{23}+\dfrac{1}{21}+\dfrac{1}{19}\right)=0\)
\(\Leftrightarrow123-x=0\Leftrightarrow x=123\)
Vậy x = 123
\(\sqrt{\dfrac{169}{64}}=\sqrt{\dfrac{13^2}{8^2}}=\dfrac{13}{8}\)
\(\dfrac{\sqrt{169}}{\sqrt{64}}=\dfrac{\sqrt{13^2}}{\sqrt{8^2}}=\dfrac{13}{8}\)
Vậy \(\sqrt{\dfrac{169}{64}}=\dfrac{\sqrt{169}}{\sqrt{64}}\)
Tương tự
Bị đao không hai căn bậc bằng nhau hết mà tính làm gì nhìn vô là biết bằng roy :V
a) \(\dfrac{x}{5}=\dfrac{y}{6};\dfrac{y}{8}=\dfrac{z}{7}\)và \(x+y-z=69\)
Theo đề bài, ta có:
\(\dfrac{x}{5}=\dfrac{y}{6}\Rightarrow\dfrac{x}{5}\times\dfrac{1}{8}=\dfrac{y}{6}\times\dfrac{1}{8}\Rightarrow\dfrac{x}{40}=\dfrac{y}{48}\)(1)
\(\dfrac{y}{8}=\dfrac{z}{7}\Rightarrow\dfrac{y}{8}\times\dfrac{1}{6}=\dfrac{z}{7}\times\dfrac{1}{6}\Rightarrow\dfrac{y}{48}=\dfrac{z}{42}\)(2)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\Rightarrow\dfrac{x}{40}=\dfrac{y}{48}=\dfrac{z}{42}=\dfrac{x+y-z}{40+48-42}=\dfrac{69}{46}=\dfrac{3}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{40}=\dfrac{3}{2}\Rightarrow x=\dfrac{40\times3}{2}=60\\\dfrac{y}{48}=\dfrac{3}{2}\Rightarrow y=\dfrac{48\times3}{2}=72\\\dfrac{z}{42}=\dfrac{3}{2}\Rightarrow z=\dfrac{42\times3}{2}=63\end{matrix}\right.\)
Vậy \(\Rightarrow\left\{{}\begin{matrix}x=60\\y=72\\z=63\end{matrix}\right.\)
Ta có:\(\dfrac{x}{5}=\dfrac{y}{6}\Rightarrow\dfrac{x}{20}=\dfrac{y}{24}\)(Nhân 2 vế với \(\dfrac{1}{4}\))
\(\dfrac{y}{8}=\dfrac{x}{7}\Rightarrow\dfrac{y}{24}=\dfrac{z}{21}\)(Nhân 2 vế với \(\dfrac{1}{3}\))
\(\Rightarrow\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}\)và x+y-z=6
Áp dụng tính chất dãy tỉ số bằng nhau. Ta có:
\(\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}=\dfrac{x+y-z}{20+24-21}=\dfrac{69}{23}=3\)
Vì \(\dfrac{x}{20}=3\Rightarrow x=20.3=60\)
\(\dfrac{y}{24}=3\Rightarrow y=24.3=72\)
\(\dfrac{z}{21}=3\Rightarrow z=3.21=63\)
Vậy x=60; y=72; z=63
a)\(\sqrt{9.4}=\sqrt{36}=6;\sqrt{9}.\sqrt{4}=3.2=6\Rightarrow\sqrt{9.4}=\sqrt{9}.\sqrt{4}\)
b)\(\sqrt{169-144}=\sqrt{25}=5;\sqrt{169}-\sqrt{144}=13-12=1\Rightarrow\sqrt{169-144}>\sqrt{169}-\sqrt{144}\)
tra loi ho mik lun di mai ik hoc roi !chut chut chuit chut
a, 1+2y / 18 = 1+4y / 24 = 1+6y / 6x
Ta có : 1+2y / 18 = 1+6y / 6x = 1+2y + 1+6y / 18 + 6y
= 2+ 8y / 18+6y = 2 (1+4y) / 2( 9 +3y) = 1+4y/9+3y
Ta lại có : 1 + 4y/24 = 1+4y / 9+3y
=> 24=9+3y => 15=3y => y=5
Vậy y=5
Nhớ like
b, 1+3y/12 = 1+5y/5x = 1+7y/4x
Ta có : 1+3y/12 = 1+7y/4x = 1+3y+1+7y / 12 +4x
= 2 + 10y / 12 +4x = 2 (1+5y) / 2 (6+2x) = 1+5y / 6+2x
Ta lại có: 1+5y / 5x = 1+5y / 6+2x
=> 5x = 6+2x => 3x = 6 => x=2
Vậy x =2
\(4x^3-3=29\Rightarrow x^3=\dfrac{29+3}{4}=8\Rightarrow x=\sqrt[3]{8}=2\)
Thay số: \(\dfrac{x+16}{9}=\dfrac{2+16}{9}=2\)
Suy ra: \(y=\left(-16\right)\cdot2+25\Leftrightarrow y=-7\) và \(z=25\cdot2-49\Leftrightarrow z=1\)
\(A=x+2y+3z\Leftrightarrow2+\left(-14\right)+3=-9\)
\(4x^3-3=29\Rightarrow x^3=\dfrac{32}{4}=2^3\Rightarrow x=3\)
\(\dfrac{19}{9}=\dfrac{2y-2.25}{-32}=\dfrac{3z+49.3}{75}=\dfrac{2y+3z+49.3-25.2}{75-32}=\dfrac{2y+3z+97}{43}\)
\(\dfrac{\left(2y+3z+3\right)+94}{43}=\dfrac{19}{9}\) \(\Rightarrow\left(x+2y+3z\right)=\dfrac{43.19}{9}-94\)
a)vì\(\dfrac{x}{3}\)=\(\dfrac{y}{4}\)=\(\dfrac{z}{5}\)=>\(\dfrac{2x}{6}\)=\(\dfrac{3y}{12}\)=\(\dfrac{5z}{25}\)và 2x+3y+5z=86
áp dụng tính chất của dãy tỉ số bằng nhau ta có
\(\dfrac{2x}{6}\)=\(\dfrac{3y}{12}\)=\(\dfrac{5z}{25}\)=\(\dfrac{2x+3y+5z}{6+12+25}\)\(\dfrac{86}{43}\)=2
vì\(\dfrac{2x}{6}\)=2=>2x=2.6=12=>x=12:2=6
\(\dfrac{3y}{12}\)=2=>3y=12.2=24=>y=24:3=8
\(\dfrac{5z}{25}\)=2=>5z=25.2=50=>z=50:5=10
vậy x=6,y=8,z=10
vì\(\dfrac{x}{3}\)=\(\dfrac{y}{4}\)=>\(\dfrac{x}{9}\)=\(\dfrac{y}{12}\)(1)
\(\dfrac{y}{6}\)=\(\dfrac{z}{8}\)=>\(\dfrac{y}{12}\)=\(\dfrac{z}{16}\)(2)
từ (1)(2)=>\(\dfrac{x}{9}\)=\(\dfrac{y}{12}\)=\(\dfrac{z}{16}\)=>\(\dfrac{3x}{27}\)=\(\dfrac{2y}{24}\)=\(\dfrac{z}{16}\)và 3x-2y-z=13
áp dụng tính chất của dãy tỉ số bằng nhau ta có
\(\dfrac{3x}{27}\)=\(\dfrac{2y}{24}\)=\(\dfrac{z}{16}\)=\(\dfrac{3x-2y-z}{27-24-16}\)=\(\dfrac{13}{-13}\)=-1
vì\(\dfrac{3x}{27}\)=-1=>3x=-1.27=-27=>x=-27x;3=-9
\(\dfrac{2y}{24}\)=-1=>2y=-1.24=-24=>y=-24:2=-12
\(\dfrac{z}{16}\)=-1=>z=-1.16=-16
vậy...
Lời giải:
Áp dụng tính chất dãy tỉ số bằng nhau:
$\frac{3x+25}{144}=\frac{2y-169}{25}=\frac{z+144}{169}=\frac{3x+25+2y-169+z+144}{144+25+169}=\frac{(3x+2y+z)+25-169+144}{144+25+169}=\frac{1}{2}$
Suy ra:
$3x+25=144.\frac{1}{2}=72\Rightarrow x=\frac{47}{3}$
$2y-169=25.\frac{1}{2}\Rightarrow y=\frac{363}{4}$
$z+144=169.\frac{1}{2}\Rightarrow z=\frac{-119}{2}$
P/s: Lần sau bạn lưu ý ghi đầy đủ yêu cầu đề bài.