Câu hỏi của Phạm Tuấn Khoa

Question

giúp e vs ạ

Quỳnh Anh · Accepted Answer

Trả lời:

Bài II :

a, $\frac{3}{8}+\frac{1}{8}:x=\frac{3}{16}$

$\Leftrightarrow\frac{1}{8}:x=\frac{3}{16}-\frac{3}{8}$

$\Leftrightarrow\frac{1}{8}:x=-\frac{3}{16}$

$\Leftrightarrow x=\frac{1}{8}:\left(-\frac{3}{16}\right)$

$\Leftrightarrow x=-\frac{2}{3}$

b, $\left|2x+1\right|+\frac{3}{2}=3$

$\Leftrightarrow\left|2x+1\right|=\frac{3}{2}$

$\Leftrightarrow\orbr{\begin{cases}2x+1=\frac{3}{2}\2x+1=-\frac{3}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=\frac{1}{2}\2x=-\frac{5}{2}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{4}\x=-\frac{5}{4}\end{cases}}}$

c, $\frac{1}{4}x^2-3x=0$

$\Leftrightarrow x\left(\frac{1}{4}x-3\right)=0$

$\Leftrightarrow\orbr{\begin{cases}x=0\\frac{1}{4}x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\x=12\end{cases}}}$

d, $7^{x+2}+2.7^x=357$

$\Leftrightarrow7^x.7^2+2.7^x=357$

$\Leftrightarrow7^x\left(7^2+2\right)=357$

$\Leftrightarrow7^x.51=357$

$\Leftrightarrow7^x=7$

$\Leftrightarrow7^x=7^1$

$\Leftrightarrow x=1$

Bài III :

a, $\frac{x}{8}=\frac{1}{5}$

$\Leftrightarrow x=\frac{1}{5}.8=\frac{8}{5}$

b, $2x:8=0,3:0,6$

$\Leftrightarrow2x:8=0,5$

$\Leftrightarrow2x=0,5.8$

$\Leftrightarrow2x=4$

$\Leftrightarrow x=2$

c, $\frac{2x-1}{4}=\frac{3}{5}$

$\Leftrightarrow5\left(2x-1\right)=12$

$\Leftrightarrow10x-5=12$

$\Leftrightarrow10x=17$

$\Leftrightarrow x=\frac{17}{10}$

d, $\frac{x-1}{3}=\frac{x+2}{5}$

$\Leftrightarrow5\left(x-1\right)=3\left(x+2\right)$

$\Leftrightarrow5x-5=3x+6$

$\Leftrightarrow5x-3x=6+5$

$\Leftrightarrow2x=11$

$\Leftrightarrow x=\frac{11}{2}$

Bài IV :

a, $A=-\left(\frac{1}{2}x-1\right)^2-\left(x^2-4\right)^4+2021$

Ta có: $-\left(\frac{1}{2}x-1\right)^2\le0\forall x;-\left(x^2-4\right)^4\le0\forall x$

$\Leftrightarrow-\left(\frac{1}{2}x-1\right)^2-\left(x^2-4\right)^4\le0\forall x$

$\Leftrightarrow-\left(\frac{1}{2}x-1\right)^2-\left(x^2-4\right)^4+2021\le2021\forall x$

Dấu "=" xảy ra khi $\hept{\begin{cases}\frac{1}{2}x-1=0\x^2-4=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\x=\pm2\end{cases}\Rightarrow x=2}}$

Vậy GTLN của A = 2021 khi x = 2

b, Ta có : $\left(2x-3y\right)^2+\left|7y-5z\right|=0$

Mà $\left(2x-3y\right)^2\ge0\forall x,y;\left|7y-5z\right|\ge0\forall y,z$

$\Rightarrow\hept{\begin{cases}2x-3y=0\7y-5z=0\end{cases}\Leftrightarrow\hept{\begin{cases}2x=3y\7y=5z\end{cases}\Leftrightarrow}\hept{\begin{cases}\frac{x}{3}=\frac{y}{2}\\frac{y}{5}=\frac{z}{7}\end{cases}\Leftrightarrow}\hept{\begin{cases}\frac{x}{15}=\frac{y}{10}\left(1\right)\\frac{y}{10}=\frac{z}{14}\left(2\right)\end{cases}}}$

Từ (1) và (2) => $\frac{x}{15}=\frac{y}{10}=\frac{z}{14}$

Đặt $\frac{x}{15}=\frac{y}{10}=\frac{z}{14}=k\left(k\ne0\right)$

$\Rightarrow\hept{\begin{cases}x=15k\y=10k\z=14k\end{cases}\Rightarrow\hept{\begin{cases}x^2=225k^2\y^2=100k^2\z=196k^2\end{cases}}}$

Ta có : x² + y² + z² = 1024

=> 225k² + 100k² + 196k² = 1024

<=> 521k² = 1024

=> k² = $\frac{1024}{521}$

=> k = $\pm\sqrt{\frac{1024}{521}}$

$\Rightarrow\hept{\begin{cases}x=15k=15\sqrt{\frac{1024}{521}}\y=10k=10\sqrt{\frac{1024}{521}}\z=14k=14\sqrt{\frac{1024}{521}}\end{cases}}$ hoặc $\hept{\begin{cases}x=15k=15.\left(-\sqrt{\frac{1024}{521}}\right)=-15\sqrt{\frac{1024}{521}}\y=10k=10.\left(-\sqrt{\frac{1024}{521}}\right)=-10\sqrt{\frac{1024}{521}}\z=14.\left(-\sqrt{\frac{1024}{521}}\right)=-14\sqrt{\frac{1024}{521}}\end{cases}}$

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