a) \(\dfrac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\) (1)

\(\Leftrightarrow9x-7=\sqrt{\left(7x+5\right)\left(7x+5\right)}\)

\(\Leftrightarrow9x-\sqrt{\left(7x+5\right)\left(7x+5\right)}=7\)

\(\Leftrightarrow9x-\sqrt{\left(7x+5\right)^2}=7\)

\(\Leftrightarrow9x-\left|7x+5\right|=7\)

\(\Leftrightarrow\left[{}\begin{matrix}9x-\left(7x+5\right)=7\left(đk:7x+5\ge0\right)\\9x-\left[-\left(7x+5\right)\right]=7\left(đk:7x+5< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\left(đk:x\ge-\dfrac{5}{7}\right)\\x=\dfrac{1}{8}\left(đk:x< -\dfrac{5}{7}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x\in\varnothing\end{matrix}\right.\)

\(\Leftrightarrow x=6\)

Vậy tập nghiệm phương trình (1) là \(S=\left\{6\right\}\)

b) \(\sqrt{4x-20}+3\sqrt{\dfrac{x+5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\) (2)

\(\Leftrightarrow\sqrt{4\left(x-5\right)}+3\cdot\dfrac{\sqrt{x+5}}{3}-\dfrac{1}{3}\cdot\sqrt{9\left(x-5\right)}=4\)

\(\Leftrightarrow\sqrt{4}\sqrt{x-5}+\sqrt{x+5}-\dfrac{1}{3}\cdot\sqrt{9}\sqrt{x-5}=4\)

\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x+5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)

\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x+5}-\sqrt{x-5}=4\)

\(\Leftrightarrow\sqrt{x-5}+\sqrt{x+5}=4\)

\(\Leftrightarrow\sqrt{x-5}=4-\sqrt{x+5}\)

\(\Leftrightarrow x-5=\left(4-\sqrt{x+5}\right)^2\)

\(\Leftrightarrow x-5=16-8\sqrt{x+5}+x+5\)

\(\Leftrightarrow-5=16-8\sqrt{x+5}+5\)

\(\Leftrightarrow-5=21-8\sqrt{x+5}\)

\(\Leftrightarrow8\sqrt{x+5}=21+5\)

\(\Leftrightarrow8\sqrt{x+5}=26\)

\(\Leftrightarrow\sqrt{x+5}=\dfrac{13}{4}\)

\(\Leftrightarrow x+5=\dfrac{169}{16}\)

\(\Leftrightarrow x=\dfrac{169}{16}-5\)

\(\Leftrightarrow x=\dfrac{89}{16}\)

Vậy tập nghiệm phương trình (2) là \(S=\left\{\dfrac{89}{16}\right\}\)

Bài 1:

a) Ta có: AE // BC (gt)

=> \(\widehat{E_1}=\widehat{B_2}\) (so le trong)

Mà \(\widehat{B_1}=\widehat{B_2}\) (BD là tia phân giác của góc ABC)

=> \(\widehat{E_1}=\widehat{B_1}\)

=> \(\Delta BAE\) cân tại B.

b) Vì:

+) \(\widehat{B_1}=\widehat{B_2}\) (BD là tia phân giác của góc ABC)

+) \(\widehat{B_{12}}=60^o\) (gt)

=> \(\widehat{B_1}=\dfrac{\widehat{B_{12}}}{2}\Rightarrow\widehat{B_1}=\dfrac{60^o}{2}=30^o\left(=\widehat{B_2}\right)\)

Xét \(\Delta BAE\) có: \(\widehat{B}+\widehat{A_1}+\widehat{E_1}=180^o\) (định lý tổng 3 góc trong 1 tam giác)

mà \(\widehat{A_1}=\widehat{E_1}\) => \(\widehat{A_1}+\widehat{E_1}=30^o+30^o=60^o\)

do đó: \(\widehat{B}=180^o-\left(\widehat{A_1}+\widehat{E_1}\right)\)

\(\Rightarrow\widehat{B}=180^o-60^o\)

\(\Rightarrow\widehat{B}=120^o\)

mik thử làm xem

\(\dfrac{1-\sqrt{2}}{2\sqrt{3}-3\sqrt{2}}\)

\(=\dfrac{\left(1-\sqrt{2}\right)\left(2\sqrt{3}+3\sqrt{2}\right)}{-6}\)

\(=\dfrac{2\sqrt{3}+3\sqrt{2}-2\sqrt{6}-6}{-6}\)

\(=-\dfrac{2\sqrt{3}+3\sqrt{2}-2\sqrt{6}-6}{6}\)

a) \(x^2-6x+5=\left(x-1\right)\left(x-5\right)\)

b) \(x^2+5x-6=\left(x-1\right)\left(x+6\right)\)

c) \(x^2-5x+4=\left(x-1\right)\left(x-4\right)\)

a) \(x^2-6x+5=\left(x-1\right)\left(x-5\right)\)

b) \(x^2+5x-6=\left(x-1\right)\left(x+6\right)\)

c) \(x^2-5x+4=\left(x-1\right)\left(x+6\right)\)

đúng rồi bạn phải là \(\sqrt{12}-\sqrt{18}\)

có cho đoạn nào cắt k mà đòi kí hiệu

\(\left(x+1\right)\left(y+10\right)=6\)

\(\Leftrightarrow\left(x+1\right)\left(y+10\right)=1\cdot6\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x+1\right)\left(y+10\right)=1\\\left(x+1\right)\left(y+10\right)=6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=1\\y+10=1\\x+1=6\\y+10=6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0;y=-9\\x=5;y=-4\end{matrix}\right.\)

Vậy có 2 cặp số x, y thỏa mãn là (0;-9) và (5;-4)

thu gọn hết sẽ thấy sự giống nhau

a) \(M=\left(71+x\right)-\left(-24-x\right)+\left(-35-x\right)\)

\(=71+x-\left(-24\right)+x+\left(-35\right)-x\)

\(=60+x\)

b) \(x-34-\left[\left(15+x\right)-\left(23-x\right)\right]\)

\(=x-34-\left(15+x-23+x\right)\)

\(=x-34-\left(-8+2x\right)\)

\(=x-34-\left(-8\right)-2x\)

\(=-26-x\)

c) \(\left(-15+\left|x\right|\right)+\left(25-\left|x\right|\right)\)

\(=-15+\left|x\right|+25-\left|x\right|\)

\(=10\)