\(4x\cdot\left(x:2\right)-3\cdot\left(1-2x\right)=7-2\cdot\left(x+1\right)\)
các bạn giải cho mk với nhé. mk đang cần lém
Tìm x :
\(3x\cdot\left(x-2\right)-2x\cdot\left(2x-1\right)=\left(1-x\right)\cdot\left(1+x\right)\)
\(\left(5x+3\right)\cdot\left(3x-5\right)-\left(x-2\right)\cdot\left(2x+1\right)=6x\cdot\left(3x+1\right)-x^2\)
\(\left(2x-1\right)\cdot\left(2x+1\right)-3\cdot\left(x-1\right)=\left(1-4x\right)\cdot\left(1-x\right)\)
\(\left(2x^2+1\right)\cdot\left(3x^2-1\right)-\left(4x^2-3\right)\cdot\left(x^2+1\right)=x\cdot\left(2x^3+1\right)\)
GIÚP MK ĐI MAI MK PHẢI NỘP RÙI !
1> 3x(x-2)-2x(2x-1)=(1-x)(1+x)
⇔\(3x^2\)-6x-\(4x^2\)+2x=1-\(x^2\)
⇔-1\(x^2\) - 4x= 1- \(x^2\)
⇔ -1\(x^2\) -4x+ \(x^2\) = 1
⇔-4x=1
⇔ x = \(\dfrac{-1}{4}\)
Tìm x :
a, \(4x^2-\left(3x+1\right)\cdot\left(2x-1\right)=2\cdot\left(x-3\right)^2\)
b.\(\left(5x-1\right)\cdot\left(x+1\right)-\left(2x-1\right)\cdot\left(2x+1\right)=x\cdot\left(x+1\right)\)
c, \(7x^2-\left(2x-3\right)^2=1+3\cdot\left(x+2\right)^2\)
\(a,4x^2-\left(3x+1\right)\left(2x-1\right)=2\left(x-3\right)^2\)
\(\Leftrightarrow4x^2-\left(6x^2-3x+2x-1\right)=2\left(x^2-6x+9\right)\)
\(\Leftrightarrow4x^2-6x^2+x+1-2x^2+12x-18=0\)
\(\Leftrightarrow-4x^2+13x-17=0\)
\(\Leftrightarrow-4\left(x^2-\dfrac{13}{4}x+\dfrac{169}{64}\right)-\dfrac{103}{16}=0\)
\(\Leftrightarrow-4\left(x-\dfrac{13}{8}\right)^2=\dfrac{103}{16}\)
\(\Leftrightarrow\left(x-\dfrac{13}{8}\right)^2=\dfrac{-103}{64}\Rightarrow\) pt vô nghiệm
\(b,\left(5x-1\right)\left(x+1\right)-\left(2x-1\right)\left(2x+1\right)=x.\left(x+1\right)\)\(\Leftrightarrow5x^2+5x-x-1-\left(4x^2-1\right)=x^2+x\)
\(\Leftrightarrow5x^2+5x-x-1-4x^2+1-x^2-x=0\) \(\Leftrightarrow3x=0\Rightarrow x=0\)
\(c,7x^2-\left(2x-3\right)^2=1+3\left(x+2\right)^2\)
\(\Leftrightarrow7x^2-\left(4x^2-12x+9\right)=1+3\left(x^2+4x+4\right)\)
\(\Leftrightarrow7x^2-4x^2+12x-9=1+3x^2+12x+12\)\(\Leftrightarrow7x^2-4x^2+12x-9-1-3x^2-12x-12=0\)\(\Leftrightarrow-22=0\) ( vô lí)
Vậy phương trình vô nghiệm
Làm giúp mình nha
\(a,\left(x-2\right)\cdot\left(x+2\right)\cdot\left(x^2-10\right)=72\)
\(b,\left(x+1\right)\cdot\left(x+2\right)\cdot\left(x+3\right)\cdot\left(x+4\right)\cdot\left(x+5\right)=40\)
\(c,\left(2x-5\right)^2=\left(4x+7\right)^2\)
\(d,\left(2x^2+3x-1\right)^2-5\cdot\left(2x^2+3x+3\right)+24=0\)
a)\((x^2- 4).(x^2 - 10) = 72 Đặt x^2 - 7 = a(1), ta có (a+3)(a-3)=72 a^2-9=72 a^2=81 a=+-9 xét 2 trường hợp a = 9 và -9 khi thay vào (1) ta có..... tự lm nốt nha \)
b) nhóm x+1 vs x+4 và x+2 vs x+3 ta sẽ có (x2+5x+4)(x2+5x+6)(x+5)=40
câu c dễ lắm... bạn có thể làm 2 cách 1 là xét 2 th khi vế trái bằng vế phải hay trái dấu vế phải
hoặc cách 2 đưa về hiệu 2 bình phương nhá.. cách này dễ hơn
Chứng minh giá trị biểu thức không phụ thuộc x :
1, \(\left(2x+1\right)^3-\left(2x-1\right)^3-2\cdot\left(4x+3\right)^2+8\cdot\left(x+3\right)^2\)
2,\(\left(2x+1\right)^2\cdot\left(x-1\right)-2\cdot\left(x-2\right)^3+x\cdot\left(3-2x\right)\cdot\left(3+x\right)-\left(3x-3\right)^2\)
1: \(=8x^3+12x^2+6x+1-8x^3+12x^2-6x+1-2\left(4x+3\right)^2+8\left(x+3\right)^2\)
\(=24x^2+2-2\left(16x^2+24x+9\right)+8\left(x^2+6x+9\right)\)
\(=24x^2+2-32x^2-48x-18+8x^2+48x+72\)
=56
2: \(=\left(4x^2+4x+1\right)\left(x-1\right)-2\left(x^3-6x^2+12x-8\right)+x\left(3-2x\right)\left(3+x\right)-\left(3x-3\right)^2\)
\(=4x^3-3x-1-2x^3+12x^2-24x+16+x\left(9-3x-2x^2\right)-\left(3x-3\right)^2\)
\(=2x^3+12x^2-27x+15+9x-3x^2-2x^3-9x^2+18x-9\)
\(=6\)
Chứng minh biểu thức không phụ thuộc x :
1, \(\left(3x-1\right)^2-2\cdot\left(2x-3\right)\cdot\left(2x+3\right)-\left(x-3\right)^2\)
2, \(\left(3x+2\right)^3-\left(3x-2\right)^3-3\cdot\left(6x-1\right)\cdot\left(6x+1\right)\)
3, \(\left(3x-5\right)^2+3\cdot\left(x+1\right)\cdot\left(x-1\right)-\left(4x-3\right)^2+\left(2x+2\right)\cdot\left(2x+1\right)\)
bài 1: tìm x, biết
\(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot\frac{4}{10}\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\frac{30}{62}\cdot\frac{31}{64}=2^x\)
bài 2:
cho: p = \(\left(x-4\right)^{\left(x-5\right)^{\left(x-6\right)^{\left(x+5\right)}}}\)
tính p(x)=7
giúp mk vs!!!!!
mk cần gấp!!
Bài làm:
Ta có: \(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.....\frac{30}{62}.\frac{31}{64}=2^x\)
\(\Leftrightarrow\frac{1.2.3.....30.31}{2.2.2.3.2.4.....2.31.2.32}=2^x\)
\(\Leftrightarrow\frac{1}{2^{31}.2^5}=2^x\)
\(\Leftrightarrow\frac{1}{2^{36}}=2^x\)
\(\Rightarrow x=-36\)
mk cần cả giải thích
giúp mk vs!!!
Thu gọn biểu thức
1,\(\left(x-2\right)^3-\left(2x+3\right)^3-7\cdot\left(1-x\right)^3\)
2,\(\left(x+5\right)\cdot\left(x^2-5x+25\right)-\left(x-2\right)\cdot\left(x^2+2x+4\right)\)
3, \(\left(2x-3\right)\cdot\left(4x^2+6x+9\right)-\left(2x+1\right)^3\)
1: \(=x^3-6x^2+12x-8-8x^3-36x^2-54x-27+7\left(x-1\right)^3\)
\(=-7x^3-42x^2-42x-35+7x^3-21x^2+21x-7\)
\(=-63x^2-21x-42\)
2: \(=x^3+125-\left(x^3-8\right)=125+8=133\)
3: \(=8x^3-27-8x^3-12x^2-6x-1=-12x^2-6x-28\)
1, Giải hệ phương trình:
\(\hept{\begin{cases}x\cdot\left|x\right|-\left(x+10\right)\cdot\left|x+10\right|=y\cdot\left|y\right|\\y\cdot\left|y\right|-\left(y+10\right)\cdot\left|y+10\right|=z\cdot\left|z\right|\\z\cdot\left|z\right|-\left(z+10\right)\cdot\left|z+10\right|=x\cdot\left|x\right|\end{cases}}\)
Giải hộ mk nhoa mk tick cho !!!!!!!!!
giải hệ phương trình :
a) \(\hept{\begin{cases}x\cdot\left(1+y-x\right)=-2\cdot y^2-y\\x\cdot\left(\sqrt{2\cdot y}-2\right)=y\cdot\left(\sqrt{x-1}-2\right)\end{cases}}\)
b) \(\hept{\begin{cases}1+x\cdot y+\sqrt{x\cdot y}=x\\\frac{1}{x\cdot\sqrt{x}}+y\cdot\sqrt{y}=\frac{1}{\sqrt{x}}+3\cdot\sqrt{y}\end{cases}}\)
Làm hộ mk nhé mk tick cho :))))))))))