Để ngâm 1 kg mơ thì cần:

$4:3=\dfrac43$ (kg đường)

Để ngâm 2 kg mơ thì cần:

$\dfrac43\cdot2=\dfrac83$ (kg đường)

Vậy: ...

Nếu ngâm `1kg` mơ thì cần số đường là:

`4 : 3 = 4/3 (kg)`

Ngâm `2kg` mơ thì cần số đường là:

`4/3 . 2 = 8/6 (kg)`

Ta có : \(\left\{{}\begin{matrix}2x=3y\\4y=3z\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{2}\Rightarrow\dfrac{x}{9}=\dfrac{y}{6}\\\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{y}{6}=\dfrac{z}{8}\end{matrix}\right.\)

`=> x/9 =y/6 =z/8=>x/9 =y/6 = (2z)/16` và `x-y+2z=57`

ADTC dãy tỉ số bằng nhau ta có :

`x/9 =y/6 = (2z)/16 = (x-y+2z)/(9-6+16) = 57/19=3`

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{9}=3\Rightarrow x=3\cdot9=27\\\dfrac{y}{6}=3\Rightarrow y=3\cdot6=18\\\dfrac{z}{8}=3\Rightarrow z=3\cdot8=24\end{matrix}\right.\)

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 a) Ta có:

∠xOy + ∠yOz = 150⁰

∠xOy - ∠yOz = 90⁰

⇒ ∠xOy = (150⁰ + 90⁰) : 2 = 120⁰

⇒ ∠yOz = 120⁰ - 90⁰ = 30⁰

b) Ta có:

∠xOy + ∠x'Oy = 180⁰ (kề bù)

⇒ ∠x'Oy = 180⁰ - ∠xOy

= 180⁰ - 120⁰

= 60⁰

xoy=(150+90):2=120

yoz=120-90=30

√0,81.(√x - 16/25) = 9/10

0,9.(√x - 16/25) = 9/10

√x - 16/25 = 9/10 : 0,9

√x - 16/25 = 1

√x = 1 + 16/25

√x = 41/25

x = 1681/625 (nhận)

Vậy x = 1681/625

\(\sqrt{0,81}\left(\sqrt{x}-\dfrac{16}{25}\right)=\dfrac{9}{10}\left(x\ge0\right)\\ \Leftrightarrow\sqrt{\dfrac{81}{100}}.\left(\sqrt{x}-\dfrac{16}{25}\right)=\dfrac{9}{10}\\ \Leftrightarrow\dfrac{\sqrt{81}}{\sqrt{100}}.\left(\sqrt{x}-\dfrac{16}{25}\right)=\dfrac{9}{10}\\ \Leftrightarrow\dfrac{9}{10}.\left(\sqrt{x}-\dfrac{16}{25}\right)=\dfrac{9}{10}\\ \Leftrightarrow\sqrt{x}-\dfrac{16}{25}=1\\ \Leftrightarrow\sqrt{x}=\dfrac{41}{25}\\ \Leftrightarrow x=\dfrac{1681}{625}\)

Vậy x = \(\dfrac{1681}{625}\)

M H K P Q I O

a/ Xét tg MKP và tg MHQ có

\(MP=MH\left(gt\right);MK=MQ\left(gt\right)\) (1)

\(\widehat{KMP}=\widehat{HMP}+\widehat{HMK}=90^o+\widehat{HMK}\)

\(\widehat{HMQ}=\widehat{KMQ}+\widehat{HMK}=90^o+\widehat{HMK}\)

\(\Rightarrow\widehat{KMP}=\widehat{HMQ}\) (2)

Từ (1) và (2) => tg MKP = tg MHQ => PK=HQ

b/

Xét tg vuông HMP có

\(\widehat{MPH}+\widehat{MHP}=90^o\)

\(\Rightarrow\widehat{MPK}+\widehat{HPK}+\widehat{MHP}=90^o\)

Ta có

tg MKP = tg MHQ (cmt) \(\Rightarrow\widehat{MPK}=\widehat{MHQ}\)

\(\Rightarrow\widehat{MHQ}+\widehat{HPK}+\widehat{MHP}=90^o\)

\(\Rightarrow\left(\widehat{MHQ}+\widehat{MHP}\right)+\widehat{HPK}=90^o\)

\(\Rightarrow\widehat{PHQ}+\widehat{HPK}=90^o\)

Xét tg HPI có

\(\widehat{PHQ}+\widehat{HPK}=90^o\left(cmt\right)\Rightarrow\widehat{HIP}=90^o\Rightarrow PK\perp HQ\)

c/

∆ABC có:

∠A + ∠ABC + ∠C = 180⁰ (tổng ba góc trong ∆ABC)

⇒ ∠ABC = 180⁰ - (∠A + ∠C)

= 180⁰ - (90⁰ + 40⁰)

= 180⁰ - 130⁰

= 50⁰

⇒ ∠B₁ = ∠B₂ = 50⁰ : 2 = 25⁰

∆ABD có:

∠A + ∠B₁ + ∠ADB = 180⁰ (tổng ba góc trong ∆ABD)

⇒ x = ∠ADB = 180⁰ - (∠A + ∠B₁)

= 180⁰ - (90⁰ + 25⁰)

= 65⁰

⇒ ∠BDE = 90⁰ - ∠ADB

= 90⁰ - 65⁰

= 25⁰

∆BDE có:

∠B₂ + ∠BDE + ∠BED = 180⁰ (tổng ba góc trong ∆BDE)

⇒ y = ∠BED = 180⁰ - (∠B₂ + ∠BDE)

= 180⁰ - (25⁰ + 25⁰)

= 130⁰

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